STPM Physics Chapter 17 Electromagnetic Induction

STPM PHYSICS CHAPTER 17 ELECTROMAGNETIC INDUCTION 1. Magnetic flux (Scalar quantity) - Strength of magnetic field at any point on the surface. - Unit: Weber (Wb) or Tm2

ϕ

θ

= BA cos

When

Long Solenoid of N turns

∑ Δ l B∥

θ = 90o ,

=

μo I

ϕ = maximum

∑ l B∥

=

Nμo I Within solenoid:

B=

2. Magnetic flux linkage (When N turns of coil is used)

Φ

= NBA cos θ

Φ = maximum Φ = μr NBA cos

When

θ

,

θ = 90o ,

μr = relative

permeability 3. Change of magnetic flux linkage

∆ Φ = NBA cos ∆ θ = N ∆ BA cos θ = NB ∆ A cos θ 4. Magnetic field Straight wire

∑ Δ l B∥

=

At the end of solenoid:

μo ∋ ¿ l ¿

1 B= 2

5. Faraday’s law 6. Lenz’s law = When the magnetic flux = The direction of induced linked with a conductor current is such that the changes, an e.m.f. which induced current produces is directly proportional to a magnetic field to the rate of change of oppose the change in the magnetic flux linkage is magnetic flux that induced in the conductor. induced the current.

dϕ −d Φ E = −N dt = dt

μo I

∑ (2 πd ) B∥

μo ∋ ¿ l ¿

=

μo I B=

μo I 2 πd

Cicular loop of N turns

∑ Δ l B∥

=

7. Rate of change of thermal energy

dE dt

μo I

∑ (2r )B ∥ N μo I B=

μo ∋ ¿ 2r ¿

=

= P = I2R

8. Induced e.m.f. in a linear conductor (N=1) - No current initially. Use Fleming’s right hand rule.

dϕ E = −N dt = −d ( BA cos θ) dt = −d ( Blx cos θ) dt

STPM PHYSICS CHAPTER 17 ELECTROMAGNETIC INDUCTION = −Blv θ

13. Back-e.m.f. in self-induction - Induced e.m.f. by a change in the current through the circuit

sin

9. Induced e.m.f. in a rotating coil (with N turns)

−N

E=

dϕ dt

θ=ωt Angular velocity,

ω

=

−N

d (BA cos θ) dt

¿2π

=

−d (NBA cos ωt ) dt =

=

ω=2 πf

( r . p60. m . )

Peak e.m.f. = Eo

−NBAω sin ωt −E o sin ωt

10. Rootmean-square e.m.f.

14. Self-inductance, L - Unit : Henry (H)

Erms = Eo √2

−E L = dI dt

11. Instantaneous power - Powermagnetic field = Fmagnetic x velocity -

Powerelectric field =

=

back −e . m . f . rate of change of current ∈the conductor

v2 R

12. Eddy current - Induced current from magnetic field that flow in closed loop in large conducting plate, produces heat loss of energy - Lareg conducting plate has low resistance, hence large I induced. - Large I and B formed against Bcoil, produces oscillation damping ( Lenz’s law) - To reduce eddy current, use metal plate with cut slots ( increase resistance ) or laminate the metal plate (avoid the formation of magnetic field)

-

15. Self-induction Phenomenon which e.m.f. is induced in a conductor when the current in the conductor changes E= =

N dϕ

=

L dI

∫ N dϕ

=

∫ L dI

ϕ

=

LI

N

−L

16. Self-induction in solenoid and coil Solenoid

Eback =

−L

dI dt

LI

=

LI

=

LI

=

N ϕ NBA

N N( μo l I ¿

=

N ϕ

=

LI

=

NBA

=

N 2 A μo 2r

μo ∋ ¿ 2r A LI = N( ¿

A

N A μo l

Coil

LI

2

L

dI dt

dϕ −N dt

L

-

STPM PHYSICS CHAPTER 17 ELECTROMAGNETIC INDUCTION 17. Energy stored in an inductor / Magnetic energy A (i) At the moment an electric motor is switched on, a large Energy is stored in the magnetic field. P = IV

dU dt =

dU

=

I L (dI)

∫ dU= U

dI dt

IL

∫ L I dI 1 2

=

LI2

18. Mutual Induction - Phenomenon which e.m.f. is induced in a conductor when the current in the neighbouring conductor changes 19. Mutual Inductance - Unit : Henry (H)

−E M = dI dt

=

e . m. f . induced∈the secondary coil rate of change of current ∈the primary coil 20. Mutual induction in solenoid M = N ϕ I M = N2 BA I N2 ( M I

μo

=

N1 I¿ l1 A

M

=

N2

21. Theory of mutual induction

μo

N1 l1

A

When current In coil 2, I flows form A to B to produce S pole starts to flow to oppose S pole from coil 1 due to sudden in coil 1 increase of I Constant current I flows from B to A follows magnetic flux in coil 1 direction of coil 1 When current is I flows from B to A to produce N–pole to attract switched S pole from coil 1 due to sudden decrease of I off in coil 1

22. Conceptual question:

amount of current flows in the armature and causes damage. Why does this happen? (ii) The electric motor is usually connected in series to an inductor to avoid the above situation from occurring. How can the inductor help to avoid the damage? Answer: (i)Some universal motor has high start torque, and the moment the motor is switched on, (in rush) a large amount of I will flow in armature and causes damage. (ii)Inductor is connected in series with motor circuit to counter the large changes of I , therefore slowly raise the I instead of ‘inrush’flow.

STPM PHYSICS CHAPTER 17 ELECTROMAGNETIC INDUCTION B. Without magnetic field B, swinging of copper plate has longer oscillation before it stops. Explain. At the metal plate swinging entering B, there is induced magnetic field from the plate that oppose / repel Boriginal . Braking force is formed to slow down the swing. Eddy current is produced, thus energy is lost.

Why there is braking force when metal plate swing in magnetic field.

Entering Plate enter magnetic field Charges of magnetic field increases According to Lenz Law, B is induced to both side of plate will repel Boriginal. Braking force is applied.

Leaving Plate leaves magnetic field. Changes of B decreases B induced in both side of plate will attract Borigin Braking force is applied.

C. Explain why rotating plane of coil produces induced emf in this form. According to induced emf = NBA sin changes in change the area exposed to magnetic field. From sin

will

ωt , the graph will formed a sine wave

D. A standard resistor made of eureka wire is non-inductively-wound as shown in figure. Explain ‘non-inductively wound’ and explain how it is achieved in the standard resistor shown. Non inductive = magnetic field is not transferred / no magnetic field. Since the eureka wire loop is made of a single wire, the I entering the side will be leaving the other side. Since the two wires are placed closed together as pair and both I flow in opposite direction, magnetic field from wires will cancel each other. Hence, no magnetic field is induced in the resistor and therefore it is non-inductively wound.

E. The figure shows the cicuit breaker of a water beaker. The circuit breaker cuts off the power supply when there is a short circuit. The circuit consists of a transformer with two primary coils X and Y and a secondary coil that is connected to an electromagnetic switch which is normally closed. The coils X and Y have the same number of turns and are wound in opposite directions. The water beaker and the primary coils are connected in series with the main supply. Suppose that at a particular moment, the live wire is positive with respect to the neutral wire. i) Draw a diagram to show the magnetic field in the transformer core due to the current in the primary coil X and Y. Hence, explain why there is no current induced in the secondary coil, and the switch S remains closed. Answer:

In coil X and Y, both magnetic field lines is as diagram : The B from coil X is cancelling B from coil Y since they are travelling in opposite direction. No B exist in secondary coil to induce current. Hence no I is induced in secondary coil and the switch remains closed. ii) If coil Y is cut and the wire is connected. What will happen? The switch S is opened as there is current induced in the secondary coil due to existence of B from coil X. Since coil Y is taken out there is no B induced. Hence, B from coil X will induce I in secondary coil. Magnetic field is formed and attract switch, thus the circuit open. No I will flow to water heater