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MAGNETIC EFFECT OF CURRENT 1. Concept of Magnetic Field
If we place or suspend a small magnetic needle near a bar-magnet, the needle rests in a definite direction if we place this needle at some other point; it rests in some other direction. This shows that the magnetic needle near the bar-magnet
je e. in
experiences a torque which turns the needle to a define direction. The region near a magnet, where a magnetic needle experiences a torque and rests in a definite direction, is called ‘magnetic field.’ The line drawn from the south to the north pole of a magnetic needle freely-suspended at a point in the magnetic field
ac k
iit
is the direction of the field at that point.
Earth is also a magnet and has a magnetic field. This is why a freely-suspended magnetic needle always rests in the north-south direction. The north pole of the
cr
needle points towards north and the South Pole towards south. This shows that the earth’s magnetic field acts from south to north.
2. Oersted’s Experiment
Oersted, in 1820, found experimentally that a magnetic field is established around a current-carrying conductor just as it occurs around a magnet. His experiment. Conducting wire AB is connected to the poles of a battery through a key. The wire is kept above a magnetic needle parallel to it (in north-south
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direction). So long there is no current in the wire, the magnetic needle remains parallel to the wire. As soon as the key is pressed, a current flows through the wire and the needle is deflected if the current is reversed, the needle is deflected in the opposite direction the deflection of the needle indicates that with the passage of current in the wire, a magnetic field is established around it. On increasing the current in the wire, or on bringing the needle closer to the wire, the deflection of the needle increases. This experiment shows that the magnetic field is produced due to the electric current. Since electric current is ‘charge in
surrounding space.
je e. in
motion’, it is concluded that moving charges produced magnetic field in the
iit
(A charge, whether stationary or in motion, produces an electric field around it. If
field.)
ac k
it is in motion, then, in addition to the electric field, it also produces a magnetic
cr
3. Force on a Charge moving in a Magnetic Field: Definition of B
A charge moving in a magnetic field experiences a deflecting force. In fact, if a charge moving through a point experience a deflecting force, then a magnetic field is said to exist at that point. This field is denoted by a vector quantity B, called the ‘magnetic field’ or ‘magnetic induction’. B is defined in terms of the force experienced by the charge moving in the field.
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If a particle carrying a positive charge q and moving with velocity v through a point p in a magnetic field experiences a deflecting for F then the magnetic field B at point P is defined by the equation F=qv x B ….(i)
The force F on the charged particle is perpendicular to the plane formed by v and B. Its sense (the way it points along the direction line) is determined by righthand screw rule of vector product. When the moving charge is positive, the direction of F is the direction of advance of a right-hand screw whose axis is
je e. in
perpendicular to the plane formed by V and B and which is rotated from V to B. The direction of F will be opposite to the direction of advance of the screw if the
iit
charge is negative but moving in the same direction.
The direction of the force on the charged particle moving in magnetic field can
ac k
also be determined by Fleming’s left-hand rule, taking ‘direction of motion of positive charge’ instead of ‘direction of current’. (This rule is given ahead in this
cr
chapter.)
The magnitude of the force on the charge particle is F=qvB sin, Where
……(ii)
is the angle between v and B the force is maximum when v and B are at
right angles,
= 900 (Fig4) It is then given by Fmax=quB …….(iii)
If v is parallel to B ( =0 or 1800), then F=0. This means that if the charged particle is moving parallel to the magnetic field, it does not experience any force. In fact, this defines the direction of magnetic field B.
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If a charged particle moving through a point P in magnetic field B does not experience a deflecting force, then the direction of motion of the particle is the direction of B at Point P.
Again, in eq. (ii) if u=0, then F=0. This means that if the charged particle is stationary in the magnetic field, then it does not experience any force. (It is important to note that a ‘stationery’ charged particle in an electric field does
je e. in
experience a force.)
The Force experienced by a moving charged particle in a magnetic field can be verified by an electron-tube. At one end of this tube is a filament, heated by
iit
electric current, and at the other end is a fluorescent screen. Hot filament emits electrons (negative charge) which fall on the screen in the form of a thin beam
ac k
and so a bright spot is observed on the screen. When this tube is placed in a magnetic field, the spot is displaced to one side. This shows that a force is acting
cr
on the electrons due to which the spot has been displaced from its initial position. If we place the electron tube in different directions in the magnetic field, we see that the displacement of the spot changes with direction and becomes zero in one particular direction. In this direction, no force acts on the electrons. This is the direction of the magnetic field.
Lorentz Force: if a particle carrying a charge q is moving in space where both an electric field E and a magnetic field B are Present, then the force on the particle will be given by:
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F=q(E+v B) This is called ‘Lorentz force’.
Unit of Magnetic Field B: The magnitude of the magnetic field is given by
B=
B=
=1
iit
Because coulomb-sec-1 – ampere.
=900, then
je e. in
If F = 1 newton, q=1 coulomb, u=1 meter/sec and
ac k
Thus, the SI unit of magnetic field B is newton/(ampere-meter), also written as N/(A-m). it is also known as ‘tesla’ (T)
cr
1T = 1 N A-1 M-1
Thus, if a charge of 1 coulomb moving with a velocity of 1 meter/sec perpendicular to a uniform magnetic field experiences a force of 1 newton, then the magnitude of the field is 1 tesla. Another SI unit of magnetic field is ‘weber/meter2’ Thus
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1 T = 1 N A -1M-1 =1 Wbm-2
In the CGS system, the magnetic field is expressed in ‘gauss’ 1 tesla = 104 gauss.
The earth’s magnetic field is about 0.5 gauss. The electromagnets used in the laboratories produce magnetic fields of the order of 1 tesla. Fields of the order of 200-400 tesla can be produced by special devices for a very short spell of time
je e. in
( 10-6 sec). It is believed that much stronger magnetic field are found in some stars which are called neutron-stars.
=
cr
=kg-sec-2 –amp-1
ac k
iit
Dimensions of B : The unit of B is
Dimension of B are [MT-2A-1]
6.1 MAGNETIC FIELD OF MOVING CHARGE
We know that a point charge q, at rest in the observer’s inertial frame, produces an electric field along the radius vector and it given by
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E=
1
4 %!
0
q r r3
If the charge is moving relative to the observer’s inertial frame, it produces a magnetic field in addition to electric field. The magnitude of which is proportional to the speed of the charge relative to the observer provided (v > a,
ac k
B = n 0 2M3 …………………………………………………….(3) 4r x
B=
n 0i 2a
cr
From equation (1), we can get field at the centre of loop, x = 0
If there are N turns in the loop
B=
n 0 Ni …………………………………………………….(4) 2a
NOTE:
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1.
Equation (3) and (4) are derived for circular loop, but they can be used for
other shape of the loop also which has symmetry about the axis of the loop.
2.
In equation (4), N may be a whole number or fraction or decimal number.
3.
For the following:
je e. in
N=( i ) 2r
4.
ac k
iit
n 0( i )i 2r B= 2a
The magnetic field at a point not on the axis of the loop is mathematically
difficult to calculate. The field lines for the circular loop are not circles, but they
cr
are closed curves that link the conductor.
Magnetic field due to a circular loop
Field between two similar coaxial circular loops
Let us consider two loops, each having N turns and carrying current i are placed at a distance 2d apart
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Assuming the current is following in the same direction in each coil, the magnetic field at a short distance x from midway point O
n 0 Nia2 1 1 2 (a2 + x1 2) 3/ 2 (a2 + x22) 3/ 2 n 0 Nia2 1 1 B= 2 (a2 + (d + x2)) 3/ 2 (a2 + (d - x2)) 3/ 2
iit je e. in
B=
The field will be uniform between the loops, if
= 0 i.e.,
ac k
n 0 Nia2 (d + x) 1 +3 (- 3) 2 2 5/ 2 = 0 2 6 + x) 2) 5/ 2 6 (a2 + (d - x)@ (a + (d@ ) 2 5/ 2 2 5/ 2 Or Now, (d + x) a + (d - x) = (d - x) a + (d + x) ............... (i) 6 @ 62 @ 5/ 2 2 5/ 2 2 2 = a + d + x + 2xd a + (d + x) 2
cr
2
2
Since x is small, so neglecting x2, we have
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6 @ = a2 + d2 + 2xd 5/ 2 5/ 2 ; E = (a2 + d2) 5/ 2 1 + 22xd 2 (a + d ) ; E = (a2 + d2) 5/ 2 1 + 25xd 2 (a + d ) = 22xd 2 0 Nia 8 2 2
cr
H 1 `a j2 B3/ 2 a + 2
n B= 8 0 Ni …………………………………………….(2) 5 5a
Variation of magnetic field between the loops.
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EXAMPLE : A wire carrying current I has the configuration shown in fig.6.16 two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc, or central angle , along the circumference of the circle, with all sections lying in the same plane. What must
be in order for B to be zero at
the centre of the circle?
SOLUTION: The field due to the straight parts of the wire ; E B1 = 2 n i 0
je e. in
4r R
ac k
iit
Out of the plane
And field due to curved part of the wire
B2 =
ci m i 2r , into the plane of the wire. 2R
cr
n0
The resultant field at the centre to be zero
B1=B2
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Or
i 6 n 0 @ n 0 2r i = = 2 i = 2rad 2 4r 2R
rad Ans
SOLENIOD A solenoid is a wire would in a closely spaced over a hollow cylindrical non-conducting core. The wire coated with an insulating material so that
je e. in
the adjacent turns physically touch each other, but they are electrically insulated. dN = ndx
If n is the number of turns per unit length, each carrying a current i,
iit
uniformly wound round a cylinder of radius a, then the number of turns
element.
0 ndx i
3 /2
cr
dB =
ac k
in length dx is ndx. Thus the magnetic field at the axial point P due to the
2 a2 x2
The direction of magnetic field is along the axis of the solenoid and in the sense of advance of a right handed screw. From geometry, we have x = a cot 180 acot and dx = a cosec2 d and hence dB =
0nisin d 2 For more Study Material and Question Bank visit www.crackiitjee.in
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Total field
ni B= 0 2
sin d
1
0ni cos 2 1 2 0ni cos 1 cos 2 ….(1) 2
or B = Special cases: Case-1:
je e. in
=
2
1 0, 2
ac k
Ө B = 0ni
iit
Solenoid is of infinite length and the point chosen is at the middle
Case-2:
cr
Solenoid is of infinite length and the point is at the end of the solenoid
1
, 2 2
B
0ni 2
NOTE: 1.
If the length of the solenoid is large compared with its radius, the internal field near its center is very nearly uniform and parallel to axis,
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and the external field near the center is very small. In practice we take the magnetic field inside very tightly wound long solenoid is uniform everywhere and zero outside it. 2.
If some material medium is present inside solenoid, the magnetic field inside it is B = r 0ni; where r is the relative permeability of the
TOROID OR ANCHOR RING
je e. in
medium.
It is a solenoid of small radius bent round to form a toroid. In an ideal toroid, the field is confined entirely within the core and is uniform. The
iit
value of magnetic field at any point on the mean circumferential line is
B = 0ni
ac k
given by
n=
cr
If N is the total turns in the toroid, then
N , 2R
N i 2R
= 0
N B 0 i 2R Or B =
0Ni 2R
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Where R is the mean radius of the ring. (b)
Field due to a current loop (or its segment) at its center: In case of a circular loop:
(i)
Each element is at the same distance from the center, i.e., r = R = constant.
(ii)
The angle between element dL and r is always as in a circle 2
Contribution of each element of B is in the same direction, i.e., out of
So Biot-Savart law
0 4
IdL r r3
ac k
B
iit
the page if the current is anticlockwise and into the page if clockwise.
for this situation yields: B=
cr
(iii)
je e. in
tangent and normal are always perpendicular to each other.
0 4
IdL R2
R d, But as dL =
so
B= 0 4
IR d 0 R2
i.e., B =
0 I 4 R
….(1)
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This is the required result and from this it is clear that: (1)
If the loop is semicircular i.e., ф = π, B=
0 I 4 R
…(2)
and will be out of the page for anticlockwise current while into the page for clockwise current respectively. If the loop is a full circle with N turns, i.e., 2N : B=
0 2 NI 4 R
je e. in
(2)
…(3)
Field at an axial point of a Solenoid:
ac k
(d)
iit
Field for this situation is depicted respectively.
If many turns of an insulated wire are wound around a cylinder the
cr
resulting coil is called a solenoid [Fig. 13.13 (A)]. The field at a point on the axis of a solenoid can be obtained by superposition of fields due to a large number of identical coils all having their centre on the axis of the solenoid. This if the desired result and from this it is clear that: (1)
If the solenoid is of infinite length and the point is well inside the solenoid,
2
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So, B
0 2nI 1 1 , 4
i.e., 3 0nI (2)
…(2)
If the solenoid is of infinite length and the point is near one end,
0
and 2
i.e., B
1 nI 2 0
je e. in
0 2nI 1 0 , 4
iit
…(3)
If the solenoid is of finite length and the point in on the perpendicular
ac k
(3)
So, B
bisector of its axis,
cr
So, B
0 2nI sin 4
with sin (4)
L L2 4R2
…(4)
If the solenoid is of finite length and the point is on its but near one end
0 So, B
0 2nI sin 4 For more Study Material and Question Bank visit www.crackiitjee.in
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with sin α = (5)
L
…(5)
L2 R 2
The field and its variation with distance along the axis of a solenoid
NOTE: If the solenoid had been of infinite length, there will be no ends and hence everywhere inside the solenoid the field will be uniform, equal to
nI which outside it will be zero. This actually happens in a 'toroid' 0
je e. in
which is a solenoid bent in a circular shape so that its two ends joint each other.
Question. A vertical straight conductor carries a current vertically upwards.
iit
Points P and Q lie respectively to the east and west of the conductor at
ac k
same distance from it. Is the field at P lesser or greater than at Q? Answer: As shown in Fig. 13.15 (A), for point P (situated east of the wire), the field of the current-carrying wire will be directed north while for point Q
cr
(situated west of the wire), it will be directed south and both will have
same magnitude BW
0I . 2d
However, as the horizontal component of earth's magnetic field BH is along the direction from south to north so if the effect's magnetic field is considered,
BP BW BH while BQ BW BH
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And so, in this situation, field at P(which is east of the vertical wire) will be greater than at Q. NOTE: In this situation there will be only one neutral point towards the west of the wire and if Q is the neutral point.
0 2I BH 4 d 0 2I 4 BH
je e. in
i.e., d
Question: Equal currents are flowing through two infinitely along parallel wires. Will there be a magnetic field at a point exactly halfway between the
ac k
opposite directions?
iit
wires when the currents in them are (a) in the same direction, (b) in
Answers: If the current in both the wires is in the same direction, the field at P
cr
due to A will be into the page while due to B out of the page and as the point P is equidistant from the wires and the wires carry equal currents,
BP BA BB 0 2I 2I 0 4 d 4 d
0
However, if the wires carry equal currents in opposite directions, the field at P due to both the wires will be in the same direction; so
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2I BP 2 0 0 4 d and will be into the page for the situation. If the directions of I in wires A and B are interchanged BP will still have the same magnitude but directed out of the page]. The lines of force for the given situations. Question. A steady current is set up in a network composed of wires of equal resistances as shown in Fig. 13.18 (A) and (B). What is the magnetic
je e. in
field at the centre P in each case: Answer: By symmetry of the problem for each current-carrying wire there is another wire such that it cancels the field of the first; e.g., the field at P due to wire AB is cancelled by the field of wire DC in Fig. 13.18 (A) and
iit
field of AB is cancelled by the field of wire GF in Fig. 13.18 (B). So the
ac k
resultant field at P due to the given network is zero in both the cases. Question. Three sections of a circuit, each section consisting of a wire curved
cr
along a circular arc (all with same radius) and two long straight wires that are tangential to the arc. In (A) the straight wires pass each other without touching. The sections carry equal currents. In which case is the magnitude of the magnetic field produced at the centre of curvature (a) greatest (b) least? Answer: Assuming that each section is composed of three elements, viz., a, b and c with anticlockwise current in the arc, the field due to each element at P will be out of the page. Now as elements a and c together in each case
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produce the same field at P
0I , so the field at P will be greatest 2R
in the case in which the field due to arc b is greatest and as for a circular
0I , i.e., B , and here 4R
arc, for point at its centre, B =
3 A , so the field will be greatest in B C 2 2
je e. in
case A and least in case C. Problem. Two infinitely long, thin, insulated straight wires lie in the x-y plane along the x and y-axes respectively. Each wire carries a current I, respectively in the positive x-direction and the positive y-direction.
iit
Find the locus of a point in this plane where the magnetic field is zero.
ac k
Solution: As field due to a long straight current-carrying wire at a distance d from it is given by
0 2I 4 d
cr
B=
so the field at due to currents in wires along x and y axes respectively will be:
BA
0 2I k 4 y
and BB
0 2I k 4 x
According to given problem,
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BA BB 0, i.e.,
0 1 1 2I 0 4 y x
This is possible only if
1 1 0, y x
je e. in
i.e., y = x So the locus is a straight line passing through the origin and subtending an angle of 45° with the positive x-axis, i.e., line AB.
Problem. A pair of stationary and infinitely long bent wires are placed in the x-y
iit
plane as shown in Fig. 13.21. The wires carry currents of 10 ampere
ac k
each as shown. The segments L and M are along the x-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of the magnetic induction at the
cr
origin O.
Solution: As point O is along the length of segments L and M so the field at O due to these segments will be zero. Further, as the point O is near one end of a long wire,
BR BP BQ =
0 I 4 RO
0 I 4 SO
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so BR
0 2I 4 d
[as RO = SO = d] Substituting the given data,
BR 107 4
Wb m2
je e. in
= 10
2 10 0.02
i.e., B is 10–4 T and out of the page. Ans.
Problem. A current of 1 ampere is flowing in the sides of an equilateral triangle
iit
of side 4.5 × 10–2 m. Find the magnetic field at the centroid of the
ac k
triangle.
Solution: As shown in Fig. 13.22, the field at the centroid
cr
BO BAB BBC BCA
And as in accordance with RHSR at O, all the fields are out of the page and point O is symmetrically situated with respect to wires:
BAB BBC BCA and hence,
B0 3BAB 3BAB Now as in case of current-carrying wire of finite length for a point at a distance d from the wire,
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B=
0 I sin sin 4 d
But as here
3
1 a a sin 60 , 3 2 3
BO 3
0 18 I 4 a
iit
0 I 3 2 4 a 2 2 3
je e. in
and d =
ac k
Substituting the given data,
18 1 4.5 102
cr
BO 107 5 = 4 10
Wb m2
i.e., resultant field is 4 × 10–5 T and out of the page if current in the loop is anticlockwise (and will be into the page if clockwise). Ans. Problem. Two straight infinitely long and thin parallel wires and spaced 0.1 m apart and carry a current of 10 ampere each. Find the magnetic field at a point distance 0.1 m from both wires in the two cases when the currents are in the (a) same and (b) opposite directions.
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Solution: As the point P is equidistant from both the wires A and B and the distance between the wires is equal to the distance of point P from a wire, so that BA = BB = B =
0 2I 2 10 107 4 d 0.1
= 2 × 10–5 T
je e. in
Now as lines of force in case of a current-carrying wire are circles encircling the wire so BA will perpendicular to AP while BB to BP. So (a) If the wire carry current in the same direction (say out of the page), B A and
iit
BB will have directions with angle between them. So BR = 2B cos 30 = 3
(b)
ac k
2 3 105 T along negative x-axis. Ans. If the wires carry current in opposite direction (say A out of the page
cr
while B into the page), BA and BB will have directions as shown in Fig.
2 between them. So 3
13.23 (B) with angle BR = 2B cos 60
= 2 × 10–5 along positive y-axis. Ans. Problem. Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper. The wire A carries a current of 9.6 ampere, directed into the plane of the paper. The wire B carries a current such that the
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10 m from 11
magnetic field of induction at the point P at a distance
the wire B is zero. Find (a) the magnitude and direction of the current in B, (b) the magnitude of the magnetic field of induction at the point S, and (c) the force per unit length on the wire B. Solution: (a) As the field due to a long current-carrying wire at a point distant d
B=
0 2I 4 d
BR BA BB
0 2IA 2IB 4 dA dB
ac k
=
iit
so the field at P will be
je e. in
from it and not near its ends is given by
cr
According to given problem BR = 0 So,
IB I A dB dA
i.e., IB = IA
dB dA
So, IB = 9.6
10 1 11 10 2 11
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= – 3A i.e., current in the wire B is 3 ampere and opposite to that in A, i.e., it will be out of the page. (b)
As lines of force for a current-carrying wire are circles encircling the wire, so BA will be perpendicular to AS while BB to BS and as in this problem AS2 + BS2 = AB2, i.e., AS is perpendicular to BS, BA and BB for given directions of current will be along SB and SA respectively.
je e. in
0 2IA along SB 4 dA
0 2IB along SA 4 dB
and BB
iit
So, BA =
B=
ac k
and hence,
B2A BB2
cr
[as SA is perpendicular to SB] i.e., B = 107
1/2
2 9.6 2 3.0 4 4 1.2 1.6
= 1.3 × 10–6 T Ans. (c)
As force per unit length between two parallel current-carrying wires separated by a distance d is given by
2I1I2 dF 0 dL 4 d
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and is attractive if current in the wires is in same direction otherwise repulsive, so here
dF 2 9.6 3 10 dL 2 6
= 2.88 10
N and repulsive Ans. m
Problem. Two long parallel wires carrying currents 2.5 ampere and I ampere in
je e. in
the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at distance of 5 m and 2 m respectively from a collinear point R.
An electron moving with velocity of 4 × 105 m/s along the positive x-
iit
(a)
ac k
direction experiences a force of magnitude 3.2 × 10–20 N at the point R. Find the value of I.
Find all the positions at which a third long parallel wire carrying a
cr
(b)
current of magnitude 2.5 ampere may be placed so that the magnetic induction at R is zero. Solution: (a)
As force on a charged particle in a magnetic field is given by
F q v B
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The field due to both current-carrying wires will be along negative y-axis. Now as the particle is moving along positive x-axis, the angle between
the direction of motion of charged particle and field, rad. 2 So, F = qvB sin 90 i.e., B =
F qv
je e. in
Substituting the given data,
3.2 1020 B= 1.6 1019 4 105
iit
= 5 × 10–7 T
ac k
Now, as this field is produced by wires P and Q and point R is on the same side of both,
2IP 2IQ d dQ P
cr
0 4
= 2 × 10–7 T i.e.,
2 2.5 2I 5 5 2
i.e., I = 4A Ans. (b)
If the distance of point R from third current-carrying wire is x, then BR = 0 implies
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0 2 2.5 5 107 0 4 x Now there are two possibilities: (i)
If the current in third wire is into the page, I = +2.5 A, so x = –1, i.e., the point R must be to the left of the wire, i.e., the wire must be to the right of R, i.e., at A with RA = 1 m.
(ii)
If the current in third wire is out of the page, I = –2.5 A, so x = +1, i.e., the
je e. in
point r must be to the right of the wire, i.e., the wire must be to the left of R, i.e., at B with RB = 1 m.
Problem: A very long thin strip of metal of width 'b' carries a current I along its length as shown in Fig. 13.26. Find the magnitude of magnetic field in
ac k
point.
iit
the plane of the strip at a distance 'a' from the edge nearest to the
Solution: Assuming the strip to be made up of a large number of elements
cr
parallel to its length, consider an element of length dx at a distance x from the point P. Treating the element as a long current-carrying wire, dB =
0 2I' 4 x
Now as the current in the strip of width b is I, so the current in the element of width dx will be I' =
I dx b
and hence,
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dB =
0 2I dx 4 b x
So, B = a b
a
0 2I 4 b
dx 0 2I b loge 1 Ans. x 4 b a
NOTE:
je e. in
If the point is distant from the strip, i.e., a > b.
=
0 2I b 4 b a
0 2I 4 a
ac k
So, B =
b a
iit
b b b2 loge 1 2 .... a a 2a
cr
i.e., for a distant point the strip behaves as a current carrying wire which justifies out result.
Problem. In the Bohr model of hydrogen atom, the electron circulates around the nucleus in the path of radius 5.1 × 10–11 m at a frequency of 6.8 × 1015 rev/s. Calculate the magnetic induction B at the centre of the orbit. What is the equivalent dipole moment? Solution: The revolving electron in equivalent to a circular current I = ef = 1.6 × 10–19 × 6.8 × 1015
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= 1.088 mA So the field at the centre of the path B=
0 2I 4 R
= 10
7
2 1.088 103 5.1 1011
= 13.4 Wb/m2 Ans.
moment
je e. in
Further, as a current-carrying loop is equivalent to a magnetic dipole of
M = NIS = R2I as N 1
2
1.088 103
iit
So, M = 5.1 1011
ac k
i.e., M = 8.88 × 10–24 A-m2 Ans.
Problem. A charge of 1 coulomb is placed at one end of a non-conducting rod of
cr
length 0.6 m. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency 104 π rad/s. Find the magnetic field at a point on the axis of rotation at a distance of 0.8 m from the centre of the path. Now half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the mid point of the rod with the same angular
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frequency. Calculate the magnetic field at a point on the axis at distance of 0.4 m from the centre of the rod. Solution: As revolving charge q is equivalent to a current I = qf = q
2
je e. in
104 =1 2 = 5 × 103 A
and field at a distance x from the centre on the axis of a current-carrying
iit
coil is given by
0 2 NIR 2 B= 4 R 2 x2 3 / 2
ac k
2 1 5 103 0.6
cr
So, B = 10
7
0.62 0.82
2
3 /2
= 1.13 × 10–3 T Ans. Now, if half of the charge is placed at the other end and the rod is rotated at the same frequency,
q
q
I' = f f 2 2 = qf = I
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= 5 × 103 A However, according to the given problem in the situation R' = 0.3 m and x' = 0.4 m, so, B' = 10
7
2 1 5 103 0.3 0.32 0.42
2
3 /2
= 2.26 × 10–3 T Ans.
je e. in
Problem. Two circular coils each of 100 turns are held such that one lies in the vertical plane and the other in the horizontal plane with their centres coinciding. The radii of the vertical and horizontal coils are respectively 20 cm and 30 cm. If the directions of the currents in them are such that
iit
the earth's magnetic field at the centre of the coils is exactly
dip = 30°]
ac k
neutralized, calculate the currents in each coil [H = 27.8 A/m; angle of
cr
Solution: As the field due to a current-carrying coil is along its axis, the vertical coil will produce horizontal field and horizontal coil vertical, i.e.,
0 2NVIV BH 4 RV and
0 2NHIH BV 4 RH
But as tan
BV , BH
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BV BH tan
BH 3
as 6
A Wb 4 107 2 m m
i.e., IV =
2 100 IV 4 107 27.8 0.2
0.2 2 27.8 0.1112 A Ans. 100
and, 107
2 100 IH 27.8 4 107 0.3 3
2 3 27.8 1000
cr
i.e., IH =
je e. in
7
iit
So, 10
ac k
and, 1
= 0.0963 A Ans.
Problem. A circular coil of radius 0.157 m has 50 turns. It is placed such that its axis is in the magnetic meridian. A dip needle is supported at the centre of the coil with its axis of rotation horizontal and in the plane of the coil. The angle of dip is 30° when a current flows through the coil. The angle of dip becomes 60° on reversing the current. Find the current in the coil assuming that the magnetic field due to the coil is smaller than the horizontal component of earth's magnetic field [BH = 3 × 10–5 T]
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B
Solution: An angle of dip, tan V and in the setting specified in the BH problem,
BH BH BC with BH > BC
tan 30 =
BV BH BC
and tan 60 =
iit
3 3 1 3
cr
=
BH BC tan60 BH BC tan30
ac k
i.e.,
BV BH BC
je e. in
So that,
So, BC =
=
1 B 2 H
1 3 105 2
= 1.5 × 10–5 T Now as field at the centre of a current-carrying coil is given by
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B=
0 2NI 4 R
so 107
i.e., I =
2NI 1.5 105 R
0.157 150 2 3.14 50
= 0.075 A Ans.
je e. in
Problem. Calculate the field at the centre of a semicircular wire of radius R in situations (A), (B) and (C) if the straight wire is of infinite length.
Solution: Keeping in mind that the field due to a straight wire of infinite length
iit
for a point at distance d from one of its ends is zero if the point is along
ac k
0I if the point is on a line perpendicular to its length 4d
its length and
0I and here 4R
cr
while at the centre of a semicircular coil is
BR Ba Bb Bc (A)
BR 0
=
0 I 0 4 R
0 I into the page 4 R
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BR
=
(C)
0 I 4 R
0 I 4 R
0 I 2 out of the page 4 R
BR
=
0 I 4 R
0 I 4 R
0 I I 0 4 R 4 R
0 I 2 into the page 4 R
je e. in
(B)
Problem. The wire loop PQRSP formed by joining two semi-circular wires of radii R1 and R2 carries a current I as shown in Fig. 13.31. What is the magnetic induction at the centre O and magnetic moment of the loop
ac k
iit
in cases (A) and (B)?
Solution: As the point O is along the length of the straight wires, so the field at O
(A)
cr
due to them will be zero and hence
B
0 I I 4 R 2 R1
i.e., B
1 0 1 I out of the page 4 R1 R 2
and, M NIS
1 1 R 22 R12 2 2
= 1 I
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i.e., M (B)
1 I R 22 R12 into the page 2
Following as in case (A), in this situation,
1 1 into the page R R 1 2
B
0 I 4
and,
1 I R 22 R12 into the page 2
je e. in
Problem. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arc AB of the ring subtends an angle Ө at the centre. What is the value of the magnetic field at the centre due to the current in the
iit
ring?
0 I 4 r
cr
B=
ac k
Solution: As the field due to an arc at the centre is given by
So, BO =
0 I1 I 2 0 2 4 r 4 r
But as, (VA – VB) = I1R1 = I2R2 i.e., I1
R1 R2
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= I1
L1 L2
(as R \ L) or, I2 I1
2
(as L = rф)
0 I1 I 0 1 4 r 4 r
0
je e. in
So, BO =
i.e., the field at the center of the coil is zero and is independent of Ө. Problem. A long wire is bent into the shape without cross contact at P.
iit
Determine the magnitude and direction of field at C. Now the circular
ac k
loop is rotated without distortion about its diameter perpendicular to the straight portion of the wire. Determine the field at C if the point A of the loop goes into the paper on rotation.
cr
Solution: The field at point C will be the resultant of the fields due to circular loop and wire. So in situation (A) as the field due to both, the loop and the wire at C will be out of the page, B R = BW + B C =
0 2I 4 R
i.e., BR
0 2I 4 R
0 2I 1 out of the page 4 R
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However, in situation (B), the field due to the coil will be along its axis (which is parallel to the straight wire) while due to straight wire out of the page. So BC and BW in the this situation will be at right angles to each other so that
=
B
2 W
0 2I 4 R
B2C
1 2 B
and, tan1 W BC
1
18
iit
= tan1
je e. in
BR =
ac k
FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor is placed in a magnetic field, the
cr
conductor experiences a force in a direction perpendicular to both the direction of magnetic field and the direction of current flowing in the conductor. This force is also called Lorentz force. F
i
B
i
The direction of this force can be found out either by Fleming’s left hand rule or by right hand palm rule.
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The magnetic force is F = ilB sin Ө
In vector form, F i l B Where
B = intesity of magnetic field i = current in the conductor l = length of the conductor Ө = angle between the length of conductor and direction of
je e. in
and
magnetic field. Case :
If Ө= 90° or sin Ө= 1 then F = ilB (maximum)
iit
(i)
ac k
Therefore, force will be maximum when the conductor carrying current is perpendicular to magnetic field. If Ө= 0° or sin Ө= 0,
cr
(ii)
Then F = ilB × 0 = 0
Thus, the force will be zero, when the current carrying conductor is parallel to the field.
Example 12. A straight current carrying conductor is placed in such a way that the current in the conductor flows in the direction out of the plane of the paper. The conductor is placed between two poles of two magnets, as shown. The conductor will experience a force in the direction towards.
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P R
S
S
N
Q
(A) P (B)
Q
(C)
R
(D) S Sol. The direction of magnetic field on the conductor is along SR.
je e. in
But F i l B
F i l kˆ (–B ˆi)
iit
F i l B ˆj i l B(ˆj)
ac k
Hence, the direction of magnetic force on the wire is towards Q.
cr
Hence option (B) is correct
Example 13. In the figure shown a semicircular wire loop is placed in uniform magnetic field B = 1.0 T. The plane of the loop is perpendicular to the magnetic field. Current i = 2A flows in the loop in the direction shown. Find the magnitude of the magnetic force in both the cases (a) and (b). The radius of the loop is 1.0m. ×
×
×
×
×
× i = 2A × × 1m × ×
×
×
×
×
(a)
×
×
×
× × i = 2A × ×
×
×
×
×
×
×
×
×
×
×
(b)
×
Sol. Refer figure (a) :
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It forms a closed loop and the current completes the loop. Therefore, net force on the loop in uniform field should be zero. Refer figure (b) : In this case although it forms a closed loop, but current does not complete
×
× C ×
×B
×
×
×
×
× A ×
×
×
×
×
× D ×
je e. in
the loop. Hence, net force is not zero.
FACD FAD
|Floop | 2|FAD |
iit
Floop FACD FAD 2FAD
ac k
|Floop | 2ilB sin (l = 2r = 2.0 m)
cr
= (2) (2) (2) (1) sin 90º = 8 N
Example 14. An arc of a circular loop of radius ‘R’ is kept in the horizontal plane and a constant magnetic field ‘B’ is applied in the vertical direction as shown in the figure. If the arc carries current ‘I’ then find the force in the arc. × × × × ×
× × × × ×
× I× × × ×
× × × B × × × × × 90º × × × × ×
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Sol. As we know, magnetic force on a closed loop placed in uniform magnetic field is zero. (1)
R
/2
R
(2)
B1 B2 0
B1 B2 IlB(ˆj)
je e. in
B1 Il B ˆj B1 IB( 2R)ˆj
iit
B1 2 IBR
ac k
Example. A conducting wire bent in the form of a parabola y2 = 2x carries a current i = 2 A wire is placed in a uniform magnetic field B 4kˆ Tesla. The
y(m)
cr
magnetic force on the wire is (in newton) A
2
x(m)
B
(A)
16 ˆi
(B)
32 ˆi
(C)
32 ˆi
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16 ˆi
(D)
Sol. The net magnetic force on closed loop is zero. A
B
Force on parabola + force on straight wire AB = 0 Force on the parabola = – force on straight wire AB
je e. in
Iˆj B
2 4 ˆj 4kˆ
iit
F 32 ˆi
ac k
Hence option (C) is correct.
cr
Example. A conductor of length ‘l’ and mass ‘m’ is placed along the east-west line on a table. Suddenly a certain amount of charge is passed through it and it is found to jump to a height ‘h’. The earth’s magnetic induction is B. The charge passed through the conductor is: (B is horizontal) (A)
1 Bmgh
(B)
2gh Blh
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(C)
gh Blh
(D)
m 2gh Bl
Sol. The magnetic force is F = I lB
or dq
Fdt lB
or q
Fdt mv 0 lB lB
q
cr
v 0 2gh
ac k
But 02 = v02 – 2gh
je e. in
dq lB dt
iit
or F
m 2gh lB
Hence option (D) is correct
Example. A U-shaped wire of mass m and length l is immersed with its two ends in mercury (see figure). The wire is in a homogeneous field of magnetic induction
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B. If a charge, that is, a current pulse q idt , is sent through the wire, the wire will jump up. ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
× l
×
×
×
B
i
m
Hg
Calculate, from the height ‘h’ that the wire reaches, the size of the charge or
je e. in
current pulse, assuming that the time of the current pulse is very small in comparison with the time of flight. Make use of the fact that impulse of force
equals Fdt , which equals mv. Evaluate ‘q’ for B = 0.1 Wb/m2, m = 10 gm, l = 20
iit
cm & h = 3 meters. [g = 10 m/s2]
Fdt = mv
But
cr
or I l Bdt = mv
ac k
Sol. I l B = F
1 2 mv mgh 2
v 2gh Idt
mv m 2gh lB lB
m 10 10 3 q 2gh 2 10 3 15 coulmb. lB 20 102 0.1
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Example. A metal ring of radius r = 0.5 m with its plane normal to a uniform magnetic field B of induction 0.2 T carries a current I = 100 A. The tension in newtons
(A)
100
(B)
50
(C)
25
(D)
10
je e. in
developed in the ring is:
Sol.
l of the loop.
/2
/2
Tcos /2
ac k
Tcos /2
iit
F
T
cr
T
+
B
For equilibrium,
F 2T
sin 2T T 2 2
or I∆lB = TӨ or IrӨB = TӨ T = IrB = 100 × 0.5 × 0.2 = 10 N
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Hence option (D) is correct Example: Crying a current of magnitude 2.5 ampere may be placed so the induction at R is zero. Solution: (a)
As force on a charged particle in a magnetic is given by
je e. in
F q v B
The field due to both current-carrying wires will be along ???
iit
y-axis. Now as the particle is moving along positive ???? the angle between the direction of motion of charged particle
ac k
rad. 2
cr
So, F = qvB sin 90 i.e., B =
F qv
Substituting the given data,
3.2 1020 B= 1.6 1019 4 105 = 5 × 10–7 T
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Now, as this field is produced by wires P and Q and point R on the same side of both,
0 2IP 2IQ 4 dP dQ = 5 × 10–7 T
2 2.5 2I 5 5 2
i.e., I = 4A Ans. (b)
je e. in
i.e.,
If the distance of point R from third current-carrying wire is x, then BR = 0 implies
ac k
iit
0 2 2.5 5 107 0 4 x Now there two possibilities: (i)
cr
If the current in third wire is into the page, I = + 2. 5 A, x = – 1, i.e., the point R must be to the left of the wire, i.e., the wire must be to the right of R, i.e., at A with RA = 1 m.
(ii)
If the current in third wire is out of the page, I = – 2.5 A, so x = +1, i.e., the point R must be to the right of the wire, i.e., the wire must be to the left of R, i.e., at B with RB = 1 m.
Problem:
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A very long thin strip of metal of width 'b' carries a current I along its length. Find the magnitude of magnetic field in the plane of the strip at a distance 'a' from the edge nearest to the point.
Solution: Assuming the strip to be made up of a large elements parallel to its length, consider an element of length dx at a distance x from the point
dB =
je e. in
P. Treating the a long element as a long current-carrying wire,
0 2I' 4 x
Now as the current in the strip of width b is I, so the current in the
I dx b
ac k
I' =
iit
element of width dx will be
cr
and hence, dB =
0 2I dx 4 b x
So, B =
=
0 2I ab dx 4 b a x
0 2I b loge 1 Ans. 2 b a
Note:
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If the point is distant from the strip, i.e., a >> b,
b b b2 loge 1 2 ... a a 2a So, B =
=
b a
0 2I b 2 b a
0 2I 4 a
which justifies our result. Problem:
je e. in
i.e., for a distant point the strip behaves as a current carrying wire
In the Bohr model of hydrogen atom, the electron circulates around
iit
the nucleus in the path of radius 5.1 × 10–11 m at a frequency of 6.8 ×
ac k
1015 rev/s. Calculate the magnetic induction B at the centre of the
Solution:
cr
orbit. What is the equivalent dipole moment?
The revolving electron is equivalent to a circular current I = ef = 1.6 × 10–19 × 6.8 × 1015 = 1.088 mA So the field at the centre of the path B=
0 2I 4 R
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= 10
7
2 1.088 103 5.1 1011
= 13.4 Wb/m2 Further, as a current-carrying loop is equivalent to a magnetic dipole of moment M = NIS
[as N = 1] –11 2
je e. in
= R2I
) × 1.088 × 10–3
i.e., M = 8.88 × 10–24 A-m2 Ans.
iit
Problem:
ac k
A charge of 1 coulomb is placed at one end of a non-conducting rod of length 0.6 m. The rod is rotated in a vertical plane about a horizontal
cr
axis passing through the other end of the rod with angular frequency 104
the magnetic field at a point on the axis of rotation
at a distance of 0.8 m from the centre of the path. Now half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the mid point of the rod with the same angular frequency. Calculate the magnetic field at a point on the axis at a distance of 0.4 m from the centre of the rod. Solution:
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As revolving charge q is equivalent to a curret I = qf = q
2
104 =1 2 = 5 × 103 A
0 2NIR 2 4 R 2 x2 3 / 2
7
2 1 5 103 0.6 0.62 0.82
2
3 /2
ac k
So, B = 10
iit
B=
je e. in
and field at a distance x from the current-carrying coil is given by
cr
= 1.13 × 10–3 T
Now, if half of die charge is placed at the other end and the rod is rotated at the same frequency,
q
q
I' = f f 2 2 = qf = I = 5 × 103 A However, according to the given problem in the situation R' = 0.3 m and x' = 0.4 m,
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So, B' = 10
3 7 2 1 5 10 0.3
0.32 0.42
2
3 /2
= 2.26 × 10–3 T Ans. Problem. Two circular coils each of 100 turns are held such that one lies in the vertical plane and the other in the horizontal plane with their centres
je e. in
coinciding. The radii of the vertical and horizontal coils are respectively 20 cm and 30 cm. If the directions of the currents in them are such that the earth's magnetic field at the centre of the coils is exactly neutralized, calculate the currents in each coil. [H = 27.8 A/m;
iit
angle of dip = 30°]
ac k
Solution:
As the field due to a current-carrying coil is along its axis, the vertical
cr
coil will produce horizontal field and horizontal coil vertical, i.e.,
0 2NVIV BH 4 R V and
0 2NHIH BV 4 RH
But as tan
BV , BH
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BV = BH tan
BH 3
as 6
A Wb 4 107 2 m m
So, 10
7
2 100 IV 0.2
= 4 107 27.8
je e. in
and, 1
= 0.1112 A Ans.
2 100 IH 0.3
ac k
and, 10
iit
0.2 2 27.8 100
i.e., IV =
7
27.8 3
cr
= 4 107
i.e., IH =
2 3 27.8 1000
= 0.0963 A Ans. Problem: A circular coil of radius 0.157 m has 50 turns. It is placed such that its axis is in the magnetic meridian. A dip needle is supported at the
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centre of the coil with its axis of rotation horizontal and in the plane of the coil. The angle of dip is 30° when a current flows through the coil. The angle of dip becomes 60° on reversing the current. Find the current in the coil assuming that the magnetic field due to the coil is smaller than the horizontal component of earth's magnetic field [BH = 3 × 10–5 T]. Solution:
tan
je e. in
As angle of dip,
BV B H
BH
iit
and in the setting specified in the problem,
ac k
B H ± BC
with BH > BC
BV BH BC
cr
So that, tan 30° =
and tan 60° =
i.e.,
=
BV BH BC
BH BC tan60 BH BC tan30
3 3 1 3
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So, BC =
=
1 BH 2
1 3 105 2
= 1.5 × 10–5 T Now as field at the centre of a current-carrying coil is given by
0 2NI 4 R
so 10
7
2NI 1.5 105 R
0.157 150 2 3.14 50
ac ki it
i.e., I =
je e. in
B=
= 0.075 A Ans. Problem:
cr
Calculate the field at the centre of a semi- circular wire of radius R in situations, if the straight wire is of infinite length. Solution: Keeping in mind that the field due to a straight wire of infinite length for a point at a distance d from one of its ends is zero if the point is
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0I if the point is on a line perpendicular to its 4d
along its length and
0I and here 4R
length while at the centre of a semicircular coil is
BR Ba Bb Bc
BR 0
(B)
BR
0 I 4 R
0 I 4 R
0 I 4 R
0 I 2 out of the page 4 R
ac k
=
BR
0 I 4 R
0 I I 0 4 R 4 R
cr
(C)
0 I into the page 4 R
je e. in
=
0 I 0 4 R
iit
(A)
=
0 I 2 into the page 4 R
Problem. The wire loop PQRSP formed by joining two semi-circular wires of radii R1 and R2 carries a current I. What is the magnetic induction at the centre O and magnetic moment of the loop in cases (A) and (B)?
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Solution: As the point O is along the length of the straight wires, so the field at O due to them will be zero and hence
B
0 I I 4 R2 R1
i.e., B
1 0 1 out of the page I 4 R1 R 2
je e. in
(A)
and, M NIS
1 1 R 22 R12 2 2
(B)
1 I R 22 R12 into the page 2
ac k
i.e., M
iit
= 1 I
Following as in case (A), in this situation,
1 0 1 into the page I 4 R1 R 2
cr
B
and, M
1 I R 22 R12 into the page 2
Problem. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arc AB of the ring subtends an angle at the centre.
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What is the value of the magnetic field at the centre due to the current in the ring? Solution: As the field due to an arc at the centre is given by
0 I 4 r
So, BO =
0 I1 0 I2 2 4 r 4 r
= I2R2
= I1
R1 R2
ac k
i.e., I2 = I1
iit
But as, (VA – VB) = I1R1
je e. in
B=
L1 L2
cr
as R L or, I2 = I1
2
as L r so, BO =
0 I1 I 0 1 4 r 4 r
0
i.e.,
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Note: In the situation, the current in the two semicircles will divide equally and will produce equal fields at the centre in opposite directions resulting in zero net field which verifies the correctness of the above result, i.e., the field at the centre is zer Problem: A long wire is bent into the shape without cross contact at P.
je e. in
Determine the magnitude and direction of field at C. Now the circular loop is rotated without distortion about its diameter perpendicular to the straight portion of the wire. Determine the field at C if the point A of the loop goes into the paper on rotation.
iit
Solution:
ac k
The field at point C will be the resultant of the fields due to circular loop and wire. So in situation (A) as the field due to both, the loop and
cr
the wire at C will be out of the page, B R = R W + BC =
0 2I 4 R
i.e., BR
0 2I 4 R
0 2I 1 out of the page 4 R
However, in situation (B), the field due to the coil will be along its axis (which is parallel to the straight wire) while due to straight wire out of
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the page. So BC and BW and in this situation will be at right-angles to each other so that BR =
=
B
2 W
B2C
0 2I 1 2 4 R
1
= tan1
18
Problem:
je e. in
B tan1 W BC
iit
A solenoid of length 0.4 m and diameter 0.6 m consists of a single
ac k
layer of 1000 turns of fine wire carrying a current of 5 × 10–3 ampere. Calculate the magnetic field on the axis at the middle and at the end
Solution:
cr
of the solenoid.
In case of a solenoid the field at a point on the axis as shown in Fig. 13.34 (A) is given by B=
0 2nI sin sin 4
Here, n
N L
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=
1000 0.4
= 2.5 × 103 turns/m × 103 × 5 × 10–3
So, B = 10–7 ×
i.e., B = 2.5 × 10–6 [sin + sin ]
L L2 4r2
=
0.4
4 7.2
4 0.3
2
iit
0.4
2
ac k
=
je e. in
So (a) when the point is at the middle on the axis =
cr
–6
×2×
4 7.2
= 8.7 × 10–6 T Ans.
L L2 r2 =
0.4
0.4
2
0.3
2
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=
4 5 10–6 ×
4 5
= 6.28 × 10–6 T Ans. Note:
=5
–6
T while for case (b), = 0 and =
, so B = 2.5 2
, so B 2 × 10–6
iit
T.
=
je e. in
If the solenoid had been of infinite length for case (a),
ac k
13.2. Force on a Current-Carrying Conductor in a Magnetic Field Through experiments Ampere established that when a current
cr
element* Id L is placed in a magnetic field B , it experiences a force
dF IdL B
…(1)
From this expression it is clear that: (a)
The magnitude of force is …(2) where is the angle between the vectors I d L and, B. So:
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(i)
Force on a current element will be minimum (= 0) when sin = min = 0, i.e., = 0° or 180°, i.e., a current element in a magnetic field does not experience any force if the current in it is collinear with the field.
(ii)
The force on the current element will be maximum (= BI dL) when sin = max = 1, i.e., = 909, i.e., force on a current element in a magnetic field is maximum (= BI dL) when it is
je e. in
perpendicular to the field. *Current element is defined as a vector having magnitude equal to the product of current with a small part of length of the conductor and the direction in which the current is
The direction of force is always perpendicular to the plane
ac k
(b)
iit
flowing in that part of the conductor.
containing Id L and B is same as that of cross-product of
cr
two vectors A B with A IdL. The direction of force when current element Id L and B are
perpendicular to each other can also be determined by applying either of the following rules: (i)
Fleming's Left-hand Rule: Stretch the forefinger, central finger and thumb of left hand mutually perpendicular. Then if the forefinger points in the direction of field ( B ), the central finger in the direction of
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current I, the thumb will point in the direction of force (or motion).
(ii)
Right-hand Palm Rule: Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point in the direction of field
B and thumb in the direction of current I, the normal to
je e. in
palm will point in the direction of force (or motion). Regarding the force on a current-carrying conductor in a magnetic field it is worth mentioning that:
As the force BI dL sin is not a function of position r, the
iit
(1)
ac k
magnetic force on a current element is non-central [a central force is of the form F = Kf (r) nr ]
The force d F is always perpendicular to both B and Id L
cr
(2)
though B and Id L may or may not be perpendicular to each other.
(3)
In case of current-carrying conductor in a magnetic field if the field is uniform, i.e., B = constt., *In this situation, d F and may or may not be zero.
F IdL B
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= I dL B
and as for a conductor
dL represents the vector sum of all
the length elements from initial to final point, which in accordance with the law of vector addition is equal to the length vector L ' joining initial to final point, so a currentcarrying conductor of any arbitrary shape in a uniform field
je e. in
experiences a force
F I dL B = IL ' B
...(3)
iit
where L ' is the length vector joining initial and final points
(4)
ac k
of the conductor.
If the current-carrying conductor in the form of a loop of any
cr
arbitrary shape is placed in a uniform field,
F
IdL B
= I dL B
and as for a closed loop, the vector sum of d L always zero. So, F = 0* [as
dL = 0]
…(4)
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i.e., the net magnetic force* on a current loop in a uniform magnetic field is always zero. Here it must be kept in mind that in this situation different parts of the loop may experience elemental force due to which the loop may be under tension or may experience a torque. (5)
If a current-carrying conductor is situated in a non-uniform
je e. in
field, its different elements will experience different forces; so in this situation,
FR 0 but
iit
may or may not be zero ...(5)
ac k
If the conductor is free to move, it translates with or without rotation.
The net force on a current-carrying conductor due to its own
field is zero; so if there are two long parallel current-carrying
cr
(6)
wires 1 and 2, wire-1 will be in the field of wire-2 and viceversa. So force on dL2 length of wire-2 due to field of wire-1, dF2 = I2 dL2dB1 =
0 2I1I2 dL 2 4 d
0 2I1 as B 1 4 d
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i.e.,
dF2 2I I 0 12 dL2 4 d
Same will be true for wire-1 in the field of wire-2. The direction of force in accordance with Fleming's left hand rule. So force per unit length in case of two parallel currentcarrying wires separated by a distance d is given by:
je e. in
dF 0 2I1I2 dL 4 d and the force between the wires is attractive if the current in them is in the same direction, otherwise repulsive [this is opposite to that of what happens between two charges].
iit
Note:
ac k
Through this concept the SI unit of current, ampere, is defined as the current which when passed through each of
cr
two parallel infinitely long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10–7 newton for one metre of their length.
(7)
In case of a current-carrying conductor in a magnet field if the conductor experiences a force and is free to move, work will be done and hence its kinetic energy or speed will change, i.e., W = KE
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with W = F.ds
Question. What is the force experienced by a semi-circular wire of radius R when it is carrying a current I and is placed in a uniform magnetic field of induction B?
je e. in
Answer: As force on a current element Id L in a field B is given by:
dF IdL B
iit
the force on a current-carrying conductor in a uniform field will be
ac k
F IdL B = I dL B
cr
[as I and B = constt.] and if n is a unit vector in the direction PQ,
dL PQ = 2R n
So, F = 2RI n B (A)
As in this situation n and B are parallel,
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= 0° F = 2RBI sin 0 = 0 Ans. (B)
As in this situation also B is perpendicular to n
F = 2RBI sin 90 = 2RBI out of the page Ans. As in this situation also B is perpendicular to n,
F = 2RBI
je e. in
(C)
and directed vertically down. Ans.
iit
Question VI:
ac k
A wire ABCDEF (with each side of length L) bent and carrying a current I is placed in a uniform magnetic induction B parallel to
Answer:
cr
positive y-direction. What is the force experienced by the wire?
If we join A to F by a straight conductor it will become a closed loop and as in case of closed loop in a uniform magnetic field,
F =0 So if FW is the force on the given network of wires and F ' on the wire AF,
FW F ' 0,
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i.e., FW F ' But as F ' = BIL sin 90° along negative z-axis, so
FW = BIL along z-axis Ans. Question. Two infinitely long parallel wires carry equal current in the same direction. What is the direction of the magnetic field due to one of
je e. in
(a)
the wires at any point along the other wire? What is the direction of force on one wire due to the other?
(c)
By what factor does this force change if the current each
iit
(b)
(d)
ac k
wire is doubled?
What is the direction of the magnetic field at a point
Answer:
cr
midway between the two wires?
(a)
In accordance with right hand screw rule, the field due to I1 at the position of I2 is perpendicular to the plane of the page and into it.
(b)
As similar currents attract each other, the force on I2 will be perpendicular to it and towards I1 .
(c)
As force per unit length between are parallel currentcarrying conductors is given by
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dF 0 2I1I2 dl 4 d so if I1 and I2 both are doubled, the force will increase by a factor of 4. (d)
For a point between the wires the field due to I1 will be into the page while due to I2 out of the page. So at a point midway between the wires for I1 = I2 the two fields will cancel
je e. in
each other and so there is no question of direction. Question.
Two parallel wires carrying current in the same direction attract each other while two similar charges moving parallel to each other in the
ac k
contradiction?
iit
same direction repel each other. Can you resolve this apparant
Answer:
cr
In case of two similar current-carrying wires only a magnetic force Fm =
0 2I1I2 4 r
…(1)
acts between the wires as the wires are electrically neutral. However, in case of motion of two similar charged particles moving parallel to each other in the same direction two forces, viz., magnetic and electric, act between them. The magnetic force is attractive and is given by Eq. (1) while the electric force is repulsive and is given by Coulomb's law, i.e.,
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Fe =
1 q1q2 40 r2
So the net force between them will be
F Fe Fm However, as the electric force is much stronger than the magnetic, so the net interaction between two similar charges moving parallel to
between them.* Problem.
je e. in
each other will be repulsive inspite of the attractive magnetic force
A straight wire of length 30 cm and mass 60 mg lies in a direction 30°
iit
east of north. The earth's magnetic field at this site is horizontal and
ac k
has a magnitude of 0.8 G. What current must be passed through the wire so that it may float in air? [g = 10 m/s2]
cr
Solution:
A current I is passed through the wire from end A towards B it will experience a force BIL sin vertically up and hence will float if BIL sin = mg i.e., I =
=
mg BL sin 60 106 10
0.8 10
4
30 10
2
1 2
= 50 A Ans.
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(A)
The Discovery and the Laws Oersted in 1820 by observing the deflection of a compass needle when it is brought near a current carrying conductor discovered that a current carrying conductor produces a magnetic field in the space surrounding it.
je e. in
The field is described by the vector B called magnetic induction, flux density or simply field – vector with units Wb/m2 or tesla. BIOT-SAVART LAW
0 IdL r 4 r3
…(1)
iit
B
B.dL I 0
ac k
or Ampere's circuital law
…(2)
cr
Biot-Savart law is an inverse-square law and is the magnetic analogue of Coulomb's law E
Gauss's law
1 r dq 3 while Ampere's law is analogous to 40 r
E.ds
1 q . 0
The direction of B at a point in case of linear and circular current carrying conductors is specified as follows : (a)
For linear currents :
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If the wire is grasped in the palm of the right hand with the stretched thumb pointing in the direction of the current, the fingers curl in the direction of the field (i.e., the lines of force). (b)
For circular currents : If the direction of the current coincides with the direction of the curl of the fingers of the right hand, the stretched thumb points in the direction of the field perpendicular to the plane of the paper is represented by (×)
je e. in
if into the page and by (.) if out of the page.
Problem 28 (b). A circular loop of radius R is bent along a diameter and given a shape as shown in the figure (A). One of the semicircles (KNM) lies in
iit
the x-z plane and the other one (KLM) in the y-z plane with their
ac k
centres at the origin. Current I is flowing through each of the semicircles.
A particle of charge q is released at the origin with a velocity 0i.
cr
(a)
Find the instantaneous force F on the particle. Assume that space is gravity free. (b)
If an external uniform magnetic field B J is applied, determine the forces
F1 and F2 on the semicircles KLM and KNM due to this field and the net force F on the loop.
Solution :
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Magnetic field due to semicircular current loop.
B
0 I . 4 R
and direction perpendicular to plane of loop outward if current is anticlockwise and towards the centre if current is clockwise if looking from that side. So, due to KNM the direction of B will be in +y-direction as looking from
je e. in
the y-axis the current is anticlockwise. Similarly due to KLM, the direction of B is along –x axis. So, due to both segments,
B BKNM BKLM
ac ki it
0I I j 0 i 4R 4R
The force on the moving charged particle in a magnetic field,
F q v B
cr
(a)
q0I 0i j i 4R
q0I k 4R 0
i.e., along (–z) direction Ans.
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(b)
The force on a current loop in external magnetic field is zero, so force F1 on segment KLM is the same as the force on segment KM carrying the current in same direction.
FKLM I dI B
I 2Rk jB 2IRBi
je e. in
and FKNM I 2Rk jB 2IRBi So, net force on the loop = 4IRB i
i.e., 4IRB Newton in + x-direction.
iit
Problem 29 (a). A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure (C). the circuit consists of
ac k
eight alternating arcs of radii r1 0.08m and r2 0.12m. Each are subtends the same angle at the centre. Find a magnetic field produced by this circuit at the centre.
(b)
An infinitely long straight wire carrying a current of 10A is passing
cr
(a)
through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the centre?
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Solution : Contribution to magnetic field due to straight wire sections converging towards centre is zero. wire sections along arcs contribute to resultant magnetic field. Magnetic field due to an arc subtending an angle at centre is given by
B
0i 2R 2
[as each arc subtends
0i 1 0i 2r1 8 4r1
je e. in
Magnetic field due to the arcs of radius r1 4
angle at centre and magnetic field due to each 4
arc is in same direction]
ac k
iit
Similarly magnetic field due to the arcs of radius r2
0i 1 1 . 4 r1 r2
cr
Hence resultant magnetic field
0i 4r2
On substituting numerical values we get
B 6.54 105 T. Ans. (b)
(i) Resultant magnetic field at the centre points upwards normal to plane of paper. The current in the wire at centre is antiparallel to the magnetic field; hence force on wire is zero.
(ii)
Magnetic field due to a long wire circulates around it hence magnetic field is tangential to arc AC so it will not experience force.
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(iii)
Section CD lies in a variable magnetic field force. So force experienced by a small element dx at a distance x from long wire at centre = Resultant force on CD
r2
r1
0i i dx 2x
0i i dx 2x
0i2 r loge 2 2 r1
je e. in
On substituting numerical values, we get
F 8.11 106 N downwards Current Loop
iit
Magnetic Moment of a Current loop
ac k
According to magnetic effects of current, in case of a current-carrying coil for a distance axial point,
0 2NIR2 B n 4 R2 x2 3/2
cr
(A)
0 2NIR2 n 4 x3 (with x >> R) If we compare this result with the field due to a small bar magnet for a distant axial point, i.e.,
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B
0 2M 4 x3
we find that a current-carrying coil for a distant point behaves as a magnetic dipole of moment
M lS with S r2Nn
…(1)
je e. in
So the magnetic moment of a current carrying coil is defined as the product of current in the coil with the area of coil in vector form. Regarding magnetic moment of a current loop it is worth noting that: (1)
Magnetic moment of a current loop is a vector perpendicular to the
iit
plane of the loop as shown in Fig. 13.57 with dimension [L2A] and unit A-
(2)
ac k
m2.
It depends on the current in the loop and its area, but is independent of
(3)
cr
the shape of the loop, i.e., circular or rectangular etc. In case of a charged particle having charge q and moving in circle of radius R with velocity v as
I qf q
v 2R
and S R2 n
1 M IS qvRn 2
…(2)
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But an angular momentum of the charged particle in this situation is
L mvRn, so M
q L 2m
…(3)
i.e., the magnetic moment of a charged particle moving in a circle is
the charge is negative. Torque on a Current Loop
When a current-carrying coil is placed in a uniform magnetic field the net
iit
force on it is always zero. However, as its different parts experience forces in different directions so the loop may experience a torque (or
ac k
couple) depending on the orientation of the loop and the axis of rotation. For this, consider a rectangular coil in a uniform field B which is free to rotate about a vertical axis PQ and normal to the plane of the coil
cr
(B)
je e. in
q times of its angular momentum and is directed opposite of L if 2m
The arm AB and CD will experience forces B(NI)b vertically up and down respectively. These two forces together will give zero net force and zero torque (as are collinear with axis of rotation), so will have no effect on the motion of the coil. Now the forces on the arms AC and BD will be BINL in the direction out of the page and into the page respectively, resulting in zero net force, but an anticlockwise couple of value.
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with S = NLb
….(1)
Now treating the current-carrying coil as a dipole of moment M IS, Eqn. (1) can be written in vector from as
M B [with M IS NiSn ]
je e. in
…(2)
This is the required result and from this it is clear that: (1)
Torque will be minimum (= 0) when s
iit
the plane of the coil is perpendicular to magnetic field i.e., normal to the
ac k
coil is collinear with the field. (2)
the plane of the coil is parallel to the field i.e., normal to the coil is
(3)
cr
perpendicular to the field.
By analogy with electric or magnetic dipole in a field, in case of a currentcarrying coil in a field,
U M.B with F =
dU dr
and W = MB(1 – The values of U and W for different orientations of the coil in the field.
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0 M is parallel to B
0 min W 0 min U MB min Stable equilibrium (A)
90 M is to B
je e. in
MB max W MB
U 0
iit
No equilibrium (B)
0 min
cr
W 2MB max
ac k
180 M is antiparallel to B
U MB max
Unstable equilibrium (C) (4)
Instruments such as electric motor, moving coil galvanometer and tangent galvanometers etc. are based on the fact the a current-carrying coil in a uniform magnetic field experiences a torque (or couple).
(C)
Galvanometers
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(a)
Moving coil galvanometer: In case of moving coil galvanometer [Fig. 13.60(A)] the deflecting torque due to current in the coil BINS* is balanced by the restoring couple dur to elasticity of spring supporting the coil. So if c is the restoring couple per unit twist and is the deflection (i.e., rotation) of the coil.
BINS c
with KG
je e. in
i.e., I KG
c Galvanometer constant NSB
iit
i.e., in case of moving coil galvanometer, the current passing through the
Tangent galvanometer :
In case of tangent galvanometer a magnetic compass needle is placed
cr
(b)
ac k
coil is directly proportional to its deflection.
horizontally at the centre of a vertical fixed current-carrying coil whose plane is in the magnetic meridian. So if the needle in equilibrium subtends an angle with the earth's magnetic field,
M BH M BC or, MBH sin MBC sin 90 i.e., BC BH tan
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or,
0 2NI BH tan 4 R
0 2NI asB C 4 R i.e., I = K tan
with K
4 RBH . Reduction factor of the tangent 0 2N
je e. in
galvanometer, i.e., in case of a tangent galvanometer when the plane of coil is in magnetic meridian, current in the coil is directly proportional to
iit
the tangent of deflection of magnetic needle.
ac k
NOTE :
A part from these, we also have a third type of galvanometer called 'hotwire galvanometer' which is based on the heating effects of current; so
cr
in it the deflection is directly proportional to the heating effect, i.e.,
I2R or I K As this galvanometer is based on he heating effects of current, it works both in ac and dc. Question-IX: For a given length L of a wire carrying a current I, how many circular turns would produce the maximum magnetic moment and of what value?
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Answer : The magnetic moment of a loop having area S and current I is given by M = IS and for a circular coil having N turns,
S R2N so, M R2NI
…(1)
Now as the length of the wire from which the loop is formed is given to
L 2RN i.e., R L / 2N
je e. in
be L, so
…(2)
L2 42N2
ac k
M NI
iit
So substituting the value of R from Eqn. (2) in (1),
…(3)
cr
IL2 i.e., M 4N
From Eqn. (3) it is clear that M will be maximum when N = min = 1, i.e., the coil has only one turn and
Mmax
1 2 IL 4
Question-X: Under what condition is the reading of a tangent galvanometer most accurate? Answer: In case of a tangent galvanometer, as
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I K tan , i.e., dI K sec2 d So,
dI d 2d I sin cos sin2
Hence the error in the measurement will be least when
sin2 max 1,
je e. in
i.e., 2 90.
This in turn implies that the reading of a tangent galvanometer will be most accurate when the deflection in it is 45.
iit
Problem 29(b). The earth has a magnetic dipole moment of
ac k
7.8 1022 A m2. What current would have to e set up in a single turn of wire going around the earth at its magnetic equator if we wish to set up such a dipole through magnetic effects of current? Given the
cr
radius of earth to be 6.4 103 km. Solution : By definition,
M IS IR2 So, I
M 7.8 1022 R 2 6.4 106
2
6.06 108 A Ans.
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Problem 30. A charge Q is spread uniformly over an insulated ring of radius R. What is the magnetic moment of the ring if it is rotated with an angular velocity with respect to the normal axis? Solution : As the charge on element dL of the ring will be
dq
Q dL 2R
so the current due to circular motion of this charge,
Q dL 2R 2
je e. in
dI dq f
as 2f
Q dL R2 2 4 R
ac k
dM dI S
iit
and hence magnetic moment due to current dI,
cr
as S R2 so, M
QR QR dL 2R 4 4
as dL 2R i.e., M
1 QR 2 Ans. 2
Problem 31. A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane
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between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5 102 T. Find the couple acting on the coil when a current of 0.1 ampere is passed through it and the magnetic field is parallel to its plane. Solution: As the coil is in the form of an equilateral triangle, its area,
1 L L sin60 2
1 3 2 0.02 2 2
3 104 m2
je e. in
S
iit
And so its magnetic moment
ac k
M IS 0.1 3 104 3 105 A m2
…(1)
cr
Now the couple on a current-carrying coil in a magnetic field is given by
MB
i.e., MB sin and when the plane of the coil is parallel to the magnetic field, the angle between M (i.e., normal to the area of coil) and B will be 90° and hence
MB sin 90 MB i.e.,
3 105 5 102
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5 3 107N m Ans. Problem 32. A solenoid of length 0.4 m a having 500 turns of wire carries a current of 3 ampere. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 ampere. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. Solution : The field of the solenoid (for an internal point not near its ends) will be
0 N 4nI 0 4 I 4 4 L
je e. in
B
iit
N asn L
500 3 0.4
ac k
i.e., B 107 4
cr
1.5 103 T
And the magnetic moment of the coil
M IS INR2
i.e., M 0.4 10 0.01
2
i.e., M 4 104 A m2 So the torque on the coil
M B MB sin
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Now as the axis of coil, i.e., direction of M is perpendicular to the axis of the solenoid, i.e., direction of B, 90 and hence
4 104 1.5 103 sin90 6 106N m Ans.
13.4 (A) Force on a Charge in a Magnetic Field
je e. in
A current element I dL in a magnetic field B experiences a force
dF IdL B
…(1)
Now if the current element I dL is due to motion of charged particles,
iit
each having a charge q moving with velocity v through a cross-section S,
ac k
I dL nS dL qv nd qv
cr
with d S dL
so Eqn. (1) reduce to
dF nd q v B
And as n d represents the total number of charged particles in volume
d (n being the number of charged particles per unit volume), the force on a charged particle will be
F
1 dF q v B n d
…(2)
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This is the required result and from this it is clear that: (1)
The force F is always perpendicular to both the velocity v and the field
B in accordance with RHSR, though v and B themselves may or may not be perpendicular to each other. (2)
If the charged particle is at rest (i.e., placed or situated) in a magnetic field, v 0, so no force will act on it, i.e., a charged particle at rest in a
(3)
je e. in
steady magnetic field does not experience any force. If the motion of the charged particle is collinear with the field, i.e.,
0 or 180,
F qB sin 0
ac ki it
[as sin 0 = sin 180 = 0]
i.e., a moving charged particle does not experience any force in a magnetic field if its motion is parallel or antiparallel to the field. As the magnitude of the force experienced by a charged particle in a
cr
(4)
magnetic field.
F qB sin , the force will be maximum qB when
sin max 1,i.e., 90,
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i.e., the particle is moving perpendicular to the field. In this situation all the three vectors F, v andB are mutually perpendicular to each other (5)
In case of motion of a charged particle in a magnetic field, as force (if it acts) is always perpendicular to the motion, i.e., displacement, so the work done,
W F.ds Fds cos 90 0 as 90 , i.e., work done in
je e. in
motion of a charged particle in a magnetic field is always zero. And as by
work-energy theorem W KE, the kinetic energy
1 m2 and 2
iit
hence speed remains constant. However, in this situation the force
particle. (6)
ac k
changes the direction of motion, i.e., the velocity v of the charged
As in case of motion of a charged particle in a magnetic field
cr
F q v B so magnetic induction B can be defined* as a vector having the direction in which a moving charged particle doesnot experience any force in the field and magnitude equal to the ratio of the magnitude of maximum force to the product of magnitude of charge with velocity, i.e. B = fmax/qv.
-particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the
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inward normal to the plane of the paper. The tracks of the particles. Relate the tracks to the particles. Answer: As force on a charged particle in a magnetic field is given by
F q v B so: (a)
As for neutron q = 0, F = 0, hence it will pass undeflected, i.e., track C corresponds to neutron. Now if the particle is negatively charged, i.e., electron, in accordance
je e. in
(b)
with RHSR, it will experience a force to the right; so track D corresponds to electron.
Further if the charged particle is positive, in accordance with RHBR, it will
-particles). Now in case of
ac k
iit
experience a force to the left; so both tracks A and B correspond to
motion of charged particle perpendicular to the magnetic field the path is a circle with radius
r
mv qB
cr
(c)
i.e., r
m q
m
4m
and as q 2e
m
m
, while q p e
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m
m
i.e., , so q q p
r rp -particle while A to proton. Question X: What is the ratio of radii of paths when an electron and a proton enter at right angles to a uniform field with same (a) velocity (b)
mp
je e. in
1840. momentum and (c) kinetic energy? Given that me Answer: In case of circular motion of a charged particle in a uniform magnetic field.
iit
2mK qB
As in this situation v, B an q are same for the two particles, so
rp re
mp
1840
cr
(a)
mv p qB qB
ac k
r
me
i.e., rp 1840 re Ans. (b)
As in this situation p, B and q are same for the two particles, so
rp re
1 1 1
i.e., rp = re Ans.
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(c)
As in this situation kinetic energy K, B and q are same for the two particle, so,
rp re
mp me
1840
i.e., rp 43 re Ans. Question XII: A beam of electrons passes undeflected through a region. What
je e. in
possible conclusions could be drawn regarding the existence of an electric and a magnetic fields?
Answer: If an electron passes undeflected through a region, the following are the probabilities of existence of an electric or a magnetic or both fields: No electric field is present and the particle is moving parallel to the
iit
(a)
(c)
ac k
magnetic field. In this situation E 0 but B 0. Both the fields are present and collinear and the particle is moving
cr
parallel to them. (This situation is equivalent to (a) + (b)) In this situation E 0 and B 0. (d)
Both the fields are present and perpendicular to each other and the particle is moving perpendicular to both of them with Fe = Fm. In this situation also E 0 and B 0.
NOTE: If the velocity of a particle remains unchanged in passing through a certain region, apart from E 0 and B 0 only situations (a), i.e.,
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E 0 and B 0 and (d), i.e., E 0 and B 0 are possible, as in other situations the speed and hence the velocity will not remain constant. Question XIII. If a charged particle is deflected either by an electric or a magnetic field, how can we ascertain the nature of the field? Answer: By observing the trajectory and measuring the kinetic energy of the charged particle as in a magnetic field the trajectory is a circle in a plane
je e. in
perpendicular to the field and the KE (and hence the speed) remains constant while in an electric field the trajectory is a parabola in a plane parallel to the field and the KE (and hence the speed) of the particle changes.
iit
Question XIV. A metallic block carrying current I is subjected to a uniform
ac k
magnetic induction B . What is the force F experienced by the charge moving with speed v and as a result of motion of this charge which face
cr
will acquire lower potential?
Answer: As the block is of metal, the charge carriers are electrons; so for current along positive x-axis, the electrons are moving along negative x-axis, i.e.
v v i.
and as the magnetic field is along the y-axis,
i.e., B B j, so F q v B
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for this case yields
F e v i B j i.e., F evBk
as i j k Ans. As force on electrons is towards the face ABCD, the electrons will
je e. in
accumulate on it and hence it will acquire lower potential*. *If the charge carriers had been positive, the face ABCD will acquire higher potential. NOTE:
iit
E.H. Hall in 1879 observed that when a current passes through a slab of
ac k
material in the presence of a transverse magnetic field, a small potential difference is established in a direction perpendicular to both, the current
cr
flow and the magnetic field. This effect is called Hall effect and the voltage thus developed is called Hall voltage. Theory shows that
VH RH
IB d
with RH
1 ne
Question XV. Can a magnetic field set a resting electron into motion? Answer: Yes; e.g., if a bar magnet is moved rapidly past a stationary charge, the charge will set into motion. this is because in this situation relative to the
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magnet, the charge will be in motion (in a direction opposite to that of the magnet) and hence it will experience a force perpendicular to the plane containing the field and the motion in accordance with RHSR. NOTE: Actually in this situation the field is not steady, and a time-varying magnetic field produces an electric field due to which the charge experiences a force and sets in motion.
je e. in
Problem 41. A uniform magnetic field with a slit system is to be used as a momentum filter for high energy charged particles. With a field of B -particle each of energy 5.3
MeV. The magnetic field is increased to 2.3B tesla and deuterons are
iit
passed into the filter. What is that energy of each deuteron
ac k
transmitted by the filter?
Solution: In case of circular motion of a charged particle in a magnetic field, as
mv qB
2mK qB
cr
r=
r2q2B2 , i.e., as K = 2m so according to the given problem,
r2 2e B2 K 2 4m 2
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r2 e 2.3B and, KD = 2 2m 2
2
K 2.3 4 i.e., D K 22 2 2
or KD = 5.3
5.29 2
je e. in
= 14.02 MeV Ans. Problem 42. A cyclotron is operating with a flux density of 3 Wb/m2. The ion which enters the field is a proton having mass 1.67 × 10–27 kg. If the maximum radius of the orbit of the particle is 0.5 m, find (a) the
iit
maximum velocity of the proton, (b) the kinetic energy of the particle,
ac k
and (c) the period for a half cycle. Solution:
As in case of motion of a charged particle in a magnetic field, r=
cr
(a)
mv qB
i.e., v =
qBr m
So, vmax =
qBrmax m
1.6 1019 3 0.5 So, vmax = 1.67 1027
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= 1.43 × 108 m/s Ans. (b)
1 mv2 2
KE =
=
1 1.67 1027 1.43 108 2
2
i.e., KE = 1.71 × 10–11 J
In case of circular motion, as T=
2 0.5 2.18 108 s, 8 1.43 10
iit
=
2 r v
ac k
(c)
je e. in
1.71 1011 108 MeV Ans. = 1.6 1019
so time for half cycle,
=
1 T 2
cr
t
1 2.18 108 1.09 108 s Ans. 2
Problem 43. A beam of charged particles, having kinetic energy 103 eV and containing masses 8 × 10–27 kg and 1.6 × 10–26 kg, emerges from the end of an accelerator tube. There is a plate at a distance 10 –2 m from the end of the tube and placed perpendicular to the beam. Calculate the
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magnitude of the smallest magnetic field which can prevent the beam from striking the plate. Solution: The motion of a charged particle in a transverse field is a circle. So with reference to the particle will not hit the plate if
mv d qB
mv as r qB 2mK d qB
ac k
or,
iit
i.e.,
je e. in
r
2mK will be maximum when m is maximum, so qd
Now as
B>
2mmaxK qd
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2 1.6 10 10 26
i.e., B >
or B >
3
1.6 1019
q 102
2 1.6 1019 q
Now as in this problem charge on the particles is not specified, assuming
B>
2 n
so, Bmin
2 T n
q
je e. in
q = ne with n = 1, 2, ….,
ac k
iit
with n = 1, 2, 3.... Ans. e
-particle is accelerated by a potential difference of 104 V. Find
cr
the change in its direction of motion if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 T [given -particle = 6.4 × 10–27 kg].
Solution:
-particle is accelerated by 104 volt, its kinetic energy will be
K = (2e) (104 V) = 2 × 104 eV. Now the path of a charged particle when it enters a magnetic field at right angles is a circle with radius
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r=
mv qB
2mK qB
So here,
2 6.4 10 r=
27
2 104 1.6 1019
1/2
2 1.6 1019 0.1
= 0.2 m
je e. in
Now as in case of a circle angle between tangents at two points will be equal to the angle between normals at those points. As in a circle tangent is normal to radius at every point, the change in direction of the
ac k
d sin1 r
iit
particle as it passes the field,
0.1 30 Ans. 0.2
cr
= sin1
Problem 45. The region between x = 0 and x = L is field with uniform, steady magnetic field B0k. A particle of mass m, positive charge q and velocity v0 i travels along x-axis and enters the region of the magnetic field. Neglect the gravity throughout the question. (a)
Find the value of L if the particle emerges from the region of magnetic field with its final velocity at an angle 30° to its initial velocity.
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(b)
Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now extends up to 2.1 L.
Solution: (a)
sin 30 =
L r
where r =
mv , radius of circular path followed by charged particle in qB
= 2.1 L =
ac k
The length of magnetic field
2.1 mv 2.1 r 2qB 2
cr
(b)
mv0 2qB
iit
=
je e. in
transverse magnetic field.
which is greater than r. So, the particle will complete the semicircle in the magnetic field as shown in figure and will emerge parallel to x-axis. Thus, Velocity on emerging from magnetic field = v0i Time spent in magnetic field =
r v0
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=
mv0 qB0v0
=
m Ans. qB0
Problem 46. A particle of mass m = 1.6 × 10–27 kg and charge q = 1.6 × 10–19 coulomb enters a region of uniform magnetic field of strength 1 tesla along the direction. The speed of the particle is 107 m/s. (a) The
je e. in
magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at point F. Find the distance EF and . (b) If the direction of the magnetic field is along the outward normal t the plane of the paper, find the time spent by the
iit
particle in the region of the magnetic field after entering it at E.
When a charged particle moves perpendicular to a magnetic field the path is a circle with magnetic force qvB directed radially towards the
cr
(a)
ac k
Solution:
centre of the path. So if we draw normals at E and F they will meet at O which is the centre of the circle.
2 1 45 Ans.
as
1
2
And as here
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r=
mv qB
1.6 1027 107 0.1 m = 1.6 1019 1 So, EF = (2 × 0.1 × sin 45) m = 10 2 cm Ans. (b)
Now if the direction of the field is out of the page the force will act as
je e. in
shown in Fig. 13.87 (B) and so now O will be the centre and the angle
3
subtended by the path at O will be 2 . Now as in a 2 2
iit
magnetic field the speed of a particle remains constant, so if t is the time
ac k
taken by the particle in the field,
r v
cr
i.e., t =
3 0.1 2 So, t = 107 -8
= 4.71 × 10–8 s Ans. Problem 47. A beam of protons with a velocity 4 × 105 m/s enters a uniform magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Find
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the radius of the helical path taken by the proton beam. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation). Mass of the proton = 1.67 × 10–27 kg. Solution: When a charged particle is projected at an angle to a magnetic field, while
r=
m v sin qB
je e. in
radius.
2 102 m 3
cr
=
ac ki it
3 1.67 1027 4 105 2 = 1.6 1019 0.3
= 1.2 cm Ans.
And as T=
2r v sin
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2 1.2 102 = 3 4 105 2 = 2.175 × 1010–7 s So pitch
1 2.175 107 2
je e. in
5 = 4 10
i.e., p = 4.35 × 10–2 m = 4.35 cm Ans.
iit
Problem 48. An electron gun G emits electrons of energy 2 keV travelling in the
ac k
positive x-direction. The electrons are required to hit the spot S where GS = 0.1 m and line GS makes an angle of 60° with the x-axis as shown in Fig. 13.89. A uniform magnetic field B parallel to GS exists in the
cr
region outside the electron gun. Find the minimum value of B needed to make the electron hit S. Solution: Resolving the velocity along the perpendicular to the magnetic field B we find that the particle will travel al the same time it will move in a circle, i.e., the path of the particle will be helical with r=
mv sin qB
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So, T =
2r v sin
m qB
= 2
and p = v cos T
2m v cos qB
So the particle will hit S if GS = np
2m v cos qB
2m v cos q GS
cr
i.e., B = n
ac k
where n = 1, 2, 3, …
iit
=n
je e. in
=
And for B to be minimum, n = min = 1 i.e., Bmin =
=
2 cos mv q GS
2 cos 2mK q GS
as mv p 2mK
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1 2 2 i.e., Bmin = 0.1
2 9.1 1031 2 103 1.6 1019 1.6 1019 Bmin =
10 8 3 104 16
= 4.71 103 T Ans.
je e. in
= 15 104
Problem 49. An electron beam passes through a magnetic field of 2 × 10 –3 Wb/m2 and an electric field of 3.4 × 104 V/m, both acting
iit
simultaneously. If the path of electrons remains undeflected, calculate
ac k
the speed of the electrons. If the electric field is removed, what will be the radius of the electron path [me = 9.1 × 10–31 kg]?
cr
Solution: An electron in an electric field experiences a force eE opposite to the field; so it will pass undeflected only if the magnetic field exerts a force which is equal in magnitude and opposite in direction of the that exerted by the electric field, i.e., E, B and v are mutually perpendicular In this situation, F e = Fm, i.e., eE = evB
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E 3.4 104 so, v = B 2 103 = 1.7 × 107 m/s Ans. Now when the electric field is switched off, the particle in the magnetic field will describe a circle with radius r=
mv qB
= 4.83 × 10–2 m
iit
= 4.83 cm Ans.
je e. in
9.1 1031 1.7 107 = 1.6 1019 2 103
Problem 50. A particle of mass 1 × 10–26 kg and charge + 1.6 × 10–19 coulomb
ac k
travelling with a velocity 1.26 × 106 m/s in +x direction enters a region in which a uniform electric field E and a uniform magnetic field B are
cr
present such that Ex = Ey = 0, Ez = – 102.4 kV/m and Bx = Bz = 0, By = 8 × 10–2 Wb/m2. The particle enters this region at the origin at time t = 0. (a) Determine the location (x, y and z co-ordinates) of the particle at t = 5 × 10–6 s. (b) If the electric field is switched off at this instant with the magnetic field still present, what will be the position o the particle at t = 7.45 × 10–6 s? Solution:
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According to the given problem the electric field is along negative z-axis while the magnetic field along +y-axis and the particle having positive charge is moving along the x-axis., So the electric force,
Fe qE
while the magnetic force
FB q v B
je e. in
= 1.6 1019 102.4 103 k
iit
= 1.6 × 10–19 × 1.28 × 106 × 8 × 10–2 k
So that,
ac k
i.e., FB 1.6 1019 102.4 103 k
F Fe Fm 0
cr
(a)
and as resultant force is zero, a=
F 0 m
And hence the particle will move along +x-axis with constant velocity
v 1.28 106 i m / s. So the distance travelled by the particle in time 5 × 10–6 s, x0 = vt
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= (1.28 × 106) × (5 × 10–6) = 6.40 m and hence the position of the particle at t = 5 × 10–6 s will be [6.4, 0, 0] m. Ans. Now when the electric field is switched off, the particle will describe a circle in x-z plane with radius
=
mv qB
je e. in
r=
1 1026 1.28 106 1m 1.6 1019 8 102
iit
With reference to Fig. 13.91, the position of the particle
ac k
x = x0 y=0
and z = r(1 –
cr
(b)
ut , so r
i.e.,
1.28 106 7.45 5 106 1 = 1.28 × 2.45 = 3.136
rad
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= 6.4 m and z = 1(1 – So the co-ordinates of the point are (6.4, 0, 2m) Ans. Problem 51. There is a constant homogeneous electric field of 100 V m–1 within the region x = 0 and x = 0.167 m pointing in the positive x-direction. There is a constant homogeneous magnetic field B within the region x = 0.167 m and x = 0.334 m pointing in the z-direction. A proton at rest at
je e. in
the origin (x = 0, y = 0) is released in a positive x-direction. Find the minimum strength of the magnetic field B, so that the proton is detected back at x = 0, y = 0.167 m (mass of the proton = 1.67 × 10 –27 kg).
iit
Solution: The situation described in the problem.
ac k
As electric field is along x-axis, so proton will be accelerated by the electric field and will enter the magnetic field at A (i.e., x = 0.167, y = 0)
cr
with velocity v along x-axis such that
1 mv2 W Fd qEd 2 1/2
2qEd i.e., v = m
1/2
2 1.6 1019 100 0.167 = 1.67 1027
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= 4 2 104
m 2
Now as proton is moving perpendicular to magnetic field so it will describe a circular path in the magnetic field with radius r such that r=
mv qB
And as it comes back at C[x = 0, y = 0.167 m] its path in the magnetic
je e. in
field will be semicircle such that
2mv qB
i.e., B =
2mv qy
i.e., B =
2 1.67 1027 4 2 104 1.6 1019 0.167
ac k
1 102 7.07 mT Ans. 2
cr
=
iit
y = 2r =
NOTE: To solve the problem properly here we have assumed that B is along negative z-axis (and not along z-axis as given in the problem). If B is along z-axis the proton will never reach the specified position of the detector. Problem 52. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields E and B are present.
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E and B are parallel to each other. At time t = 0 the velocity v0 of the particle is perpendicular to E. (Assume that, its speed is always
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