Magnetic Fields 2: Electromagnetic Induction

Question 1

Electromagnetic induction occurs whenever…
What is induced?

Answer 1

Whenever a wire cuts across the lines of a magnetic field, inducing an emf.
If the wire is part of a complete circuit, the induced emf forces e-s round the circuit (ie forces an induced current).

Question 2

Induced emf can be increased by (3)

Answer 2

-moving the wire faster
-using a stronger magnet
-making the wire into a coil, and pushing the magnet in or out of the coil.

Question 3

Other methods of generating an induced emf inc: (2)

Answer 3

Using an electric motor in reverse
Using a cycle dynamo

Question 4

Using an electric motor in reverse - how does it work (think attached weight)

Answer 4

The falling weight makes the motor coil turn between the poles of the magnet in the motor. the emf induced in the coil forces a current round the circuit. the faster the coil turns, the brighter the lamp is.

Question 5

Using a cycle dynamo - how does it work

Answer 5

When the magnet in the dynamo spins, an emf is induced in the coil, forcing a current round the circuit.

Question 6

In all these examples, an emf is induced because?

Answer 6

bc there’s a relative motion between the coil and the magnet. the induced emf is zero when the relative motion between the magnet and wires ceases.

Question 7

An electric current transfers …

Answer 7

transfers energy from the source of the emf in a circuit to the other components in the circuit. (eg energy transferred from dynamo to lamp.)

Question 8

The current though the dynamo coil causes a reaction force on… due to…

Answer 8

on the coil due to the magnet.

Question 9

Work must therefore be done to..

Answer 9

to keep the magnet spinning.

Question 10

The energy transferred from the coil to the lamp is equal to..
(assuming?)

Answer 10

to the work done on the coil to keep it spinning, assuming no energy is wasted as sound or due to friction or internal energy.

Question 11

The rate of transfer of energy from the source of emf to the other components of the circuit is equal to?
eq?

Answer 11

..equal to the product of the induced emf and the current. P=VI

This is bc:
- induced emf x current = energy transferred from source per unit charge x charge flow per second = energy transferred per second from the source.
-

Question 12

When a beam of e-s is directed across a magnetic field, each e-..

Answer 12

experiences a force (Fleming’s left hand rule/dynamo rule)

Question 13

If a rod is moved across a magnetic field, the magnetic field forces the free e-s in the rod to one end away from the other end (+ve to -ve). In this way — is induced in the rod.

Answer 13

In this way an emf is induced!

If relative motion is zero, induced emf is zero bc magnetic field no longer exerts a force on the e-s in the rod.

Question 14

What’s a solenoid?

Answer 14

A long coil.

Question 15

Lenz’s law:
Consider the north pole of a bar magnet approaching end X of a coil. The induced —- creates a —- in the —– which opposes the —–

Answer 15

The induced current creates a magnetic field in the coil which opposes the incoming N-pole.

Question 16

The induced polarity of end X must ∴ be a N-pole so as to repel the incoming N-pole ∴ current must go round X in —-direction (RHR). v.v

Answer 16

current must go round X in an acw direction (RHR). (and v.v)

Question 17

Lenz’s law states?

Answer 17

The direction of the induced emf/current is always such as to oppose the change that causes the emf/current.

(best to say emf)

Question 18

Why is Lenz’s law true?

(explain fully in terms of energy)

Answer 18

Energy is never created or destroyed. The induced current could never be in a direction to help the change that causes it, that would mean producing electrical energy from nowhere.

Leo says must lose Ek when gain Eep so slows down.

Question 19

When a conductor is moved in a magnetic field, we have F (=BIL) force to oppose its movement. So an —— is needed to keep it moving in teh field.

Answer 19

An applied force is needed.

Question 20

If the conductor moves distance Δs in time Δt:

Work done by applied force?
Charge transfer?
Hence induced emf?

Answer 20

-Work done, W, by the applied force is W = FΔs = BILΔs
-Charge transfer along conductor in this time Q = IΔt
-∴ Induced emf = W/Q =BILΔS/IΔt = BLΔs/Δt (=BLv)
-As LΔs is area ‘swept out’ by the conductor in time Δt, ε = BA/Δt

Question 21

What’s magnetic flux?

Units?

Answer 21

The product of the perpendicular magnetic flux density, B, and the area, A, swept out
Φ = BA
Unit is weber (Wb)
So induced emf is magnetic flux per second (ε = BA/Δt).

Question 22

Flux linkage through a coil of N turns is ?eq

Answer 22

NΦ = BAN, where B is the magnetic flux density perpendicular to area A.

Question 23

When the coil is turned through 180, flux linkage is ?

Answer 23

-BAN!!

Question 24

When parallel, flux linkage is?

Answer 24

Zero

Question 25

NΦ = BAN cos θ, whats θ?

Answer 25

θ is the angle between the magnetic field and the normal to the plane of the coil.

Question 26

Faraday’s law of electromagnetic induction states?

Answer 26

The induced emf in a circuit is equal to the rate of change of flux linkage through the coil.
ε = -N ΔΦ/Δt

(minus to show Lenz’s law)

Question 27

Rectangular coil of width w length L moving into uniform field at constant velocity v.

Time taken to enter field completely ?

During this time flux linkage incresases steadily from 0 to — therefore change in flux linkage per sec is —-?

When the coil is completely in the field, he flux linkage through it stays constant ∴

Answer 27

t = w/v

NΦ increases from 0 to BNLw
∴ N ΔΦ/Δt = BNLw / w/v = BNLv.

∴ induced emf is zero.

Question 28

The alternating current generator:
The simple ac generator consists of?

Answer 28

A rectangular coil that spins in a uniform magnetic field. When the coil spins at a steady rate, the flux linkage changes continuously.

Question 29

Ac generator:
For a coil spinning at a steady frequency, θ = 2πft (=ωt).
So flux linkage NΦ ( = BAN cos2πft) changes with time by the graph:

Answer 29

NΦ against t,
cosine graph,
gradient is ε,

Question 30

The induced emf is zero when..
At a max when..

Answer 30

.. when sides of coil move parallel to field lines bc NΦ = 0

.. when sides of coil cut field lines at right angles. At this point ε in each wire of each side is BLv. So for two sides, εmax = 2BLv ∴ can increase εmax by increasing v or using stronger magnet.
(—-> ε=BANωsinωt)

Question 31

For a rotating coil, the rate of change of flux is greatest flux…
rate of change of flux is zzero when flux….

Answer 31

when flux through it is zero.

when flux through it is greatest.

Question 32

Back emf:
An emf is induced in the spinning coil of an electric motor because the flux linkage through the coil changes. The induced emf, ε, is referred to as back emf bc…

Answer 32

Bc it acts against against the pd, V, applied to the motor (Lenz’s law). At any instance,
V-ε = IR, where I is the current through the motor coil and R is the circuit resistance.

Question 33

Because ε ∝ speed, v of rotation of motor, I changes as motor speed changes, ∴ …

Answer 33

∴ at low speed, current is high bc induced emf is low, v.v

Question 34

Multiply eq by I and rearrange to get…

Answer 34

IV = Iε + I^2R
Electrical power supplied by source (IV) = electrical power transferred to mechanical power (Iε) + electrical power wasted due to circuit resistance (I^2R).
∴ efficiency of electric motor = mechanical power output/ electrical power supplied.

(mechanical energy is Ek and/or Ep)

Question 35

Use an oscilloscope to display the waveform (ie variation with time) of the alternating p.d from a signal generator.
Connect the signal generator to an LED and make the f low enough so you can see the brightness of the lamp vary.

Explain brightness at diff f in relation to current.

Answer 35

At very low f, LED lights up then fades out repeatedly. Brightest twice in each cycle, corresponding to current at its peak value in each direction.
At f increases, flickering gets faster and faster until its too fast to notice.

Question 36

The heating effect of an a.c:
Imagine a heater supplied with an a.c. at very low f. The heater would heat up then cool down then heat up etc.
The heating effect P = IV = I^2R, where R is resistance of heater element.
P (=I^2R) varies with time in.. (2) (think I)?

Answer 36

At peak current, Io, max power is supplied = Io^2R
At zero current, no power is supplied.

Question 37

For a sinusoidal current, the mean power over a full cycle is half the peak power (symmetrical shape of power curve). Mean power is?

Answer 37

P = 1/2 Io^2 R

Question 38

The root mean square value of an a.c. is?

Answer 38

Irms = the value of the direct current that would give the same heating effect as the a.c. in the same resistor.

Question 39

Transformers:
What does a transformer do?

Answer 39

A transformer changes an alternating p.d to a different peak value.

Question 40

What do transformers consist of?
How does it work?

Answer 40

2 coils (primary and secondary), with the same iron core.
When the primary coil is connected to a source of alternating p.d, an alternating magnetic field is produced in the core. The field passes through the secondary coil, and an alternating emf is induced in the secondary coil by the changing magnetic field.

Question 41

A transformer is designed so that all the —- produced by the primary coil passes through the secondary coil.

Answer 41

Magnetic flux

Question 42

The induced emf in the primary coil opposes the p.d applied to the primary coil. –Therefore all the applied p.d acts against the induced emf in the primary coil. (equal)
What assumption led to this conclusion.

Answer 42

Assuming the resistance of the primary coil is negligible.

Question 43

What’s the transformers rule?

Answer 43

Vs/Vp = Ns/Np

Question 44

What’s a step-up transformer?

Answer 44

A step-up transformer has more turns on secondary coil than primary coil, so secondary voltage is stepped up compared with primary voltage.
ie Ns>Np ∴ Vs>Vp

(v.v for step-down transformer)

Question 45

The changing magnetic flux in the core induces a back emf in the primary coil as well as an emf in the secondary coil. The back emf acts against the primary voltage, making the primary current very small when the secondary current is ‘off’.
What abt when it’s ‘on’?

Answer 45

When the secondary current is on, the magnetic field it creates is in the opposite direction to the magnetic field of the primary current. In this situation, the back emf in the primary coil is reduced so the primary current is larger than when the secondary current is off.

Question 46

Transformers are almost 100% efficient bc they’re designed with: (3)

Answer 46

-low-resistance windings
-a laminated core
-a core of soft iron

Question 47

Low-resistance windings

Answer 47

To reduce power wasted due to the heating effect of the current.

Question 48

A laminated core which consists of layers of iron separated by layers of insulator. What does this do and two reasons why that’s good.

Answer 48

Induced currents in the core itself, eddy currents, are reduced in this way so the magnetic flux is as high as possible. Also, the heating effect of he induced currents in the core is reduced.

Question 49

A core of ‘soft iron’

Answer 49

It’s easily magnetised and demagnetised. This reduces power wasted through repeated magnetisation and demagnetisation of the core.

Question 50

Efficiency of transformer ?

Answer 50

Power delivered by secondary coil / power supplied to primary coil = IsVs / IpVp

Question 51

Bc almost 100% efficient, electrical power supplied to primary coil = electrical power supplied by secondary coil ∴ current ratio is?

Answer 51

Is/Ip = Vp/Vs = Np/Ns

Question 52

In a step- up transformer, the voltage is — and the current is —-

v.v

Answer 52

stepped up,
stepped down.

Question 53

The grid system:
Each power station generates a.c at f = ?, at V = ?

Stepped up to V=? for long distance transmission.

Stepped down to V=?

Answer 53

50Hz at 25kV

400kV

230V

Question 54

Transmission of electrical power over long distances is much more efficient at —- —- than at — —-

Answer 54

high voltage,
low voltage

Question 55

This is because the —- needed to deliver a certain amount of power is reduced if the voltage is increased.

Answer 55

current

So the power wasted due to the heating effect of the current through the cables is reduced.

Question 56

The power wasted through heating the cables is
I^2R = P^2R / V^2 . This means the higher the voltage….

Answer 56

the smaller the ratio of wasted power to the power transmitted is.