1. Magnetic flux (Scalar quantity)
- Strength of magnetic field at any point on the surface. - Unit: Weber (Wb) or Tm2
ϕ
= BA cos
θ
Whenθ
= 90o ,ϕ
= maximum2. Magnetic flux linkage (When N turns of coil is used)
Φ
= NBA cos
θ
Whenθ
= 90o ,Φ
= maximumΦ
=
μ
rNBA cos
θ
,μ
r = relative permeability3. Change of magnetic flux linkage
∆ Φ
= NBA cos∆ θ
= N∆
BA cosθ
= NB
∆
A cosθ
4. Magnetic field Straight wire∑
Δ l B
∥=
μ
oI
∑
(2 πd )B
∥=
μ
oI
B =
2 πd
μ
oI
Cicular loop of N turns
∑
Δ l B
∥=
μ
oI
∑
(2r )B
∥=
N μ
oI
B =
μ
o∋
¿
2 r
¿
Long Solenoid of N turns
∑
Δ l B
∥=
μ
oI
∑
l B
∥=
Nμ
oI
Within solenoid:B =
μ
o∋
¿
l
¿
At the end of solenoid:B =
1
2
μ
o∋
¿
l
¿
5. Faraday’s law= When the magnetic flux linked with a conductor changes, an e.m.f. which is directly proportional to the rate of change of magnetic flux linkage is induced in the conductor.
6. Lenz’s law
= The direction of induced current is such that the induced current produces a magnetic field to oppose the change in the magnetic flux that induced the current.
E =
−
N
dϕ
dt
=
−
dt
d Φ
7. Rate of change of thermal energy
dE
dt = P = I
2R
8. Induced e.m.f. in a linear conductor (N=1)- No current initially. Use Fleming’s right hand rule.
E = −N
dϕ
dt
=
−
d (BA cos θ)
dt
=
−
d (Blx cos θ)
dt
STPM PHYSICS CHAPTER 17 ELECTROMAGNETIC INDUCTION
= −Blv sin
θ
9. Induced e.m.f. in a rotating coil (with N turns) E =
−
N
dϕ
dt
=−
N
d (BA cos θ)
dt
=−
d (NBA cos ωt )
dt
=−
NBAω sin ωt
=−
E
osin ωt
θ=ωt
Angular velocity,ω
ω=2 πf
¿
2 π
(
r . p . m .
60
)
Peak e.m.f. = Eo 10. Root-mean-square e.m.f.E
rms=
E
o√
2
11. Instantaneous power- Powermagnetic field = Fmagnetic x velocity - Powerelectric field =
v
2R
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)
13. Back-e.m.f. in self-induction
- Induced e.m.f. by a change in the current through the circuit 14. Self-inductance, L - Unit : Henry (H)
L =
−
E
dI
dt
=
back −e . m . f .
rate of change of current∈the conductor
15. Self-induction- Phenomenon which e.m.f. is induced in a conductor when the current in the conductor changes
E =
−
N
dϕ
dt
=−
L
d I
dt
N dϕ
=L dI
∫
N dϕ
=∫
L dI
Nϕ
= LI16. Self-induction in solenoid and coil
Solenoid Coil LI = N
ϕ
LI = NBA LI = N(μ
oN
l
I
¿
A L =N
2A μ
ol
LI = Nϕ
LI = NBA LI = N(μ
o∋
¿
2 r
¿
A L =N
2A μ
o2 r
Eback =−
L
dI
dt
17. Energy stored in an inductor / Magnetic energy - Energy is stored in the magnetic field.
P = IV
d U
d t
= I Ld I
dt
dU = I L (dI)∫
dU
=∫
L I dI
U =
1
2
LI
2 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)
M =
−
E
dI
dt
=
e . m. f . induced∈the secondary coil
rate of change of current∈the primary coil
20. Mutual induction in solenoidM I = N
ϕ
M I = N2 BA M I = N2 (μ
oN
1l
1I
¿
A M = N2μ
oN
1l
1 A 21. Theory of mutual induction When currentstarts to flow in coil 1
In coil 2, I flows form A to B to produce S pole to oppose S pole from coil 1 due to sudden increase of I
Constant current in coil 1
I flows from B to A follows magnetic flux direction of coil 1
When current is switched off in coil 1
I flows from B to A to produce N–pole to attract S pole from coil 1 due to sudden decrease of I
22. Conceptual question:
A (i
) At the moment an electric motor is switched on, a large 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.
Ent e ring
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.
L e a vi n g
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 will change the area exposed to magnetic field.
From sin
ωt
, the graph will formed a sine waveD. 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.
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
