Magnetic Fields

Physics Factsheet January 2003

Number 45

Magnetic Fields Field of a long, straight wire carrying a current The lines of force of a long, straight wire carrying a current are concentric circles, in the plane at right angles to the wire, with the separation of the lines inversely proportional to the distance from the wire.

This Factsheet will explain: • what is meant by a magnetic field; • the meaning of the term magnetic flux density; • the shape of fields around wires, coils and solenoids; • the factors affecting fields around wires, coils and solenoids; • how to measure magnetic fields. Before studying the Factsheet, it is helpful to revise GCSE work on magnetic fields i.e. how to investigate the shape of a magnetic field using iron filings or a plotting compass and the shape of the field around a bar magnet; also knowledge that a magnetic field can be caused by an electric current in a coil or a solenoid. Factsheets 33 and 35 also contain related ideas about fields that would enhance understanding of this topic.

The field of a long straight wire consists of concentration circles in the plane at right angles to the wire, dropping off as r1 , where r is the distance from the wire.

Fields A field is the region around a mass, charge or magnet in which a force can be experienced. This definition is common to the three types of field – gravitational, electric and magnetic. Any of these fields can be described by “lines of force”. The line of force shows the direction of the force on a unit mass, positive charge or North magnetic pole and the density of the lines of force shows the strength of the field. Another name for “lines of force” is “flux”, so the flux density gives the strength of the field.

The direction of the lines can be remembered by using the right hand as shown below.

current field lines

The strength of the field is given by the "flux density"

Field Strength The strength of a gravitational field is defined as the force on a unit mass, and the strength of an electric field as the force on a unit positive charge. The strength of a magnetic field can be defined as the force on a unit N pole, but this is not very helpful, since unit poles do not actually exist and magnetic fields can also have an effect on charges in the field. So we shall see that magnetic fields are generally defined in terms of the force they exert on a current-carrying conductor.

Field of a coil A coil is really a long straight wire coiled into a flat circle. The field of a coil can be worked out by adding the contributions from each of the sections of the wire. North pole on opposite side

Field of Magnadur magnets A less common, but for some applications, more important shape of magnet than the bar-magnet is the “magnadur”. The shape of the field between two magnadurs is shown in below. Since the lines of force (the flux) are parallel, then the field is constant (since the density of the lines shows the field strength) for most of the centre of the region between the magnets.

current direction

field direction South pole on this side

magnadur magnet

The direction of the field - If you look at the end of the coil and the current is clockwise, then you are looking at a S pole, conversely, if the current is anti-clockwise you are looking at a N pole.

The field of two magnadurs is constant in the centre.

1

Physics Factsheet

Magnetic Fields Field of a solenoid A solenoid is a long coil (though it is often wrongly called a coil; a coil is flat.) The field is the addition of the fields from lots of coils.

Expression for the strength of a magnetic field It seems likely that the force on the current-carrying conductor should depend on the strength of the field, the length of the wire in the field and the size of the current. Each of these factors may be investigated one at a time, keeping the other factors constant, by using magnadurs mounted on a Cyoke and passing a current through a wire placed in the field. If the whole arrangement is mounted on a sensitive balance, the force can be recorded. It is indeed found that the force depends on the current I, the length l of the conductor in the field and the field strength B, so we may write:

current direction

current direction South pole on this side

F ∝ BIl

Again the direction of the field can be worked out. When you look at the end of the solenoid, the current is clockwise, you are looking at a S Pole.

In fact we use this expression to define the field strength of a magnetic field and choose the units (the tesla) so that the constant of proportionality is one. So: F = BIl

You should recognize this field as identical to that of a bar magnet, except that it continues inside the solenoid. It has the advantage over the bar magnet in that the strength of the field can be varied, whereas a bar magnet has a fixed strength.

One tesla is that field, which in a wire carrying a current of 1amp produces a force of 1 newton for each metre of the wire’s length.

Force on a current-carrying conductor in a magnetic field The force arises out of the interaction of the two magnetic fields: the one due to the wire and the one due to magnadur field. Imagine the two fields superimposed on each other. Lines of force cannot cross, since they show the direction of the force at that point in the field, so the resultant of the two forces becomes the new line of force and a combined field results. N

Exam Hint: -You will be expected to be able to use this expression linking force, flux density, current and length of the wire in the field. F = BIl

Typical Exam Question A uniform magnetic field of 40 mT acts at right angles on 3 m of a long straight wire carrying a current of 3.4 A. Calculate the force on the wire. F = BIl, therefore F = 40 × 10-3× 3.4 × 3 = 0.41N

S

=

+

magnadur field

field of a wire carrying a current away from you

You should know the definition of the tesla T, the unit of magnetic field strength (flux density) as:

combined field gives a net force downwards

That field, in which a wire carrying a current of 1 amp, produces a force of 1 N for each 1metre of the length of the wire in the field, at right angles to both the wire and the field.

Sometimes the combination results in a “neutral point” – a point where the two fields exactly cancel each other out.

Measuring magnetic fields A neutral point is a point where two or more magnetic fields cancel each other out to give no resultant force.

It is not easy to measure magnetic fields directly, since apparatus to measure the force on a wire is cumbersome. A Hall probe uses the effect of the magnetic field on electrons in a semi-conductor to give a p.d. across a slice of semi-conductor, which is proportional to the field. This gives a measure of the field strength. If an absolute value of the B-field is required the Hall probe must be calibrated in a known field, but often it is only relative values that are needed so the calibration is not always necessary.

In the case of the long straight wire in a magnadur field, there is a null point on one side of the wire, and a stronger field on the other side of the wire, so there is a net force on the wire, which is at right angles to both the magnadur field and to the wire.

Exam Hint:- Although you are expected to know about the use of a Hall probe, you are not expected to understand its internal workings.

A current-carrying conductor in a magnetic field experiences a force which is at right-angles to the field and to the wire, due to the combination of the two magnetic fields

Field of a long straight wire If the field of a long straight wire is investigated using a Hall probe it is found to depend on the current through the wire and to drop off as: 1 where r is the distance from the wire. r The constants of proportionality are such that we can write:

The direction of the force on the wire is always at right angles to the field and to the wire, but to decide the exact direction, use the diagrams above to work out which side the field due to the magnadurs, and the field due to the wire reinforce each other, or use the left-hand rule: thuMb - Motion

Forefinger - Field

µI B= 0 2π r

Where µ0 = permeability of free space (4π × 10-7 NA-2) B = magnetic field strength, flux density (Tesla, T) I = current (A) r = distance from the wire (m)

µ 0I 2 πr you will be expected to be able to carry out calculations using it Exam Hint: Although you will be given the expression B =

seCond finger - Current

2

Physics Factsheet

Magnetic Fields

Typical Exam Question Calculate the field strength at a point 60cm away from and in the plane at right angles to a long straight wire carrying a current of 1.3A µI 4π × 10-7× 1.3 B= 0 = = 4.3 × 10-7T 2πr 2π × 0.6

Exam Workshop This is a typical poor student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answers are given below. (a) Explain what is meant by field strength of a magnetic field. [3]

Field of a solenoid If the field of a solenoid is investigated using a Hall probe, it is found to be constant for most of the length of the solenoid, but drops off to about ½ its value at the ends.

It is a measure of how strong the field is.

The candidate has merely restated the question. Field strength is a precise definition. S/he should have realized that more was required for 3 marks.

It is also perhaps surprising that the field is found to be independent of the area of cross-section. It is found to be proportional to the current and to the number of turns per unit length. Again the units chosen allow the constant of proportionality to be 1. So: Where B = µ0nI

(b) Express the tesla in base units

The candidate has reduced the current and length to base units, but N is not a base unit.

n = number of turns per 1metre length µ0 = permeability of free space (4π × 10-7 NA-2) B = flux density (Tesla, T) I = current (A)

(c) Calculate the current in the cable for a household electric fire rated at 3 kW (Take mains voltage to be 240V). [2]

Exam Hint : Although you will be given the expression B = µ0nI, you should be confident in using it for calculations. Remember that n is the number of turns per 1metre length, not the total number of turns.

(d) A child stands 50cm from the cable, what will be the value of the field in the region of the child? [2] B=

You have learned that the strength of the magnetic field is the flux density, B in tesla. When you study electromagnetic induction, it is the total flux, not just its density that is important. The total flux is given the symbol Φ, and is the product of the flux density and the area.

1/2

Examiner’s Answers ✓ (a) The field strength is the flux density in Tesla, where a Tesla is that ✓ strength of field which causes a force of 1N on each 1metre length of a wire at right angles to the field,✓carrying a current of 1A. ✓-2 -1✓ (b) B is force/current × length so is kgs A (c) P = V × I therefore I = P/V = 3000/240 = 12.5 A ✓ ✓ µΙ 4π × 10-7× 12.5 = 5 × 10-6 Τ (d) B = 0 = 2πr 2π × 0.5 ✓ ✓

Φ = total flux (weber, Wb) A = cross-sectional area (m2) B = flux density (Tesla, T)

Typical Exam Question a) A solenoid is formed by winding 300 turns of wire onto a hollow cardboard tube of length 0.15m. Show that when there is a current of 0.4A in the solenoid, the magnetic flux density at the centre is 1.0 × 10-3T. [2] The solenoid has a cross-sectional area of 5 × 10-3 m2. The magnetic flux emerging from one end of the solenoid is 2.7 × 10-6 Wb. Calculate the magnetic flux density at the end of the solenoid.[2] b) Why is the flux density at the end not equal to the flux density at the centre? [1] 300 = 2000 turns/metre 0.15 B = µ0nI = 4π × 10-7× 2000 × 0.4 = 1.0 mT n=

Φ 2.7× 10-6 = = 0.54 mT A 5 × 10-3 c) Through the centre of the coil the lines of force are parallel lines so the flux density is constant, but at the ends the same lines of force spread out over a greater area, so their density is less.

b)

µ0I 4π × 10-7× 80 = = 3.2 × 10-7T 2πr 2π × 50

The candidate would not be penalized for carrying forward the error from (c), but s/he has also forgotten to change 50cm into 0.5m

Total Flux

a)

0/2

The candidate has forgotten that 3kW is 3000W and has the equation upside down. S/he should have realized that 80A is not a sensible household current.

Typical Exam Question A solenoid is 45cm long and has 100 turns. Calculate the magnetic field strength inside it when there is a current of 3.6A n = 100/0.45 = 222.2 (Remember n is turns per metre) B = 4π × 10-7 × 222.2 × 3.6T = 1.0mT

where

[2]

F =BIl, therefore B is force/current × length so units are N/Am 1/2

P =V × I therefore I = 240/3 = 80A

Φ=B×A

0/3

B=

3

Physics Factsheet

Magnetic Fields

Questions 1. Explain what is meant by a neutral point in a field. 2. (a) Sketch the magnetic field pattern of a long straight, current-carrying wire, indicating the directions of both the current and the field lines. (b) Sketch the magnetic field pattern between two magnadur magnets. Indicating the direction of the field lines. (c) Sketch the combination of the fields of (a)and (b) and explain why it results in a force on the wire which is at right angles to both the wire and the field. 3. The coil in a microphone has an average radius of 5.0 cm and consists of 250 turns. (a) Calculate the total length of the wire. The microphone magnet produces a field strength, at right angles to the coil, of 250 mT. (b) Calculate the force on the coil when it carries a current of 4mA. (c) What is the effect on the coil when the current through the coil alternates? 4. Two long straight wires, each carrying a current of 1.5 A, are placed 2 m apart in air. (a) Calculate the force which each wire exerts on each 1metre length of the other. (b) If the current in each wire is in the same direction, which direction is the force? (c) If the current in one wire is in the opposite direction to that in the other wire, which direction is the force? Answers 1. A neutral point is a point in the resultant field of two or more magnetic fields where the field stregths cancel(in magnitude and direction) out so that the net field strength is zero. 2. See text. 3. (a) Length = 2πr × 250 = 2π × 5 × 10-2 × 250 = 78.6m (b) F = BIl = 250 × 10-3 × 4 × 10-3 × 78.6 = 7.86 × 10-2N (c) The force will reverse direction each time the current changes direction, i.e. the coil will vibrate. 4. (a) Field due to one wire at the other wire µI 4π × 10-7 × 1.5 = 1.5 × 10-7T = 0 = 2πr 2π × 2 Force exerted = BIl = 1.5 × 10-7 × 1.5 × 1 = 2.25 × 10-7N (b) attracts wires (c) repels wires

Acknowledgements: This Physics Factsheet was researched and written by Janice Jones. The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

4