Physics Factsheet
www.curriculum-press.co.uk
Number 97
A2 Electromagnetic Problems
Many students find problems involving e.m. induction difficult
(compared to other Physics topics). The aim of this Factsheet is to
provide some guidance and practise in these calculations.
Exam Hint: Take care with units. the area here must be in m2.
And later on, make sure that all time references are in seconds, s.
The relevant theory will only be referred to as required by the
calculations.
Example: Find the rate of change of flux linkage when a wire
of length 25cm is pulled 80cm through a perpendicular
magnetic field of strength 0.50T over a period of 6.0 seconds.
Basic Theory
An emf is induced when a moving wire cuts through a magnetic
field, or the field strength changes through a coil. For maximum
effect, the field lines must be perpendicular to the plane of the
moving wire, or the plane of the coil.
v
25cm
Faraday said that the emf induced,
ε ∝ dΦ/dt
B = 0.50 T
Answer:
Φ = BA = 0.50 × (0.25 × 0.80) = 0.10Wb
dΦ /dt = 0.10 / 6.0 = 1.7 × 10-2Wb s-1
Lenz said that the direction of the induced current is such as to
make the induced field oppose the change causing the induction
effect. This is really a statement of Conservation of Energy, and
introduces a minus sign into the equation. In most of the calculations
demanded, this minus sign is of little importance as it just gives the
direction of the induced emf at any moment.
Example:
(a) The North Pole of a bar magnet is pushed towards the end
of a coil as shown.
B
Flux linkage Φ = BAn
v
!
EMF induced ε = -
A
n turns
B
N
dΦ
dt
A
Flux cut Φ = BA
EMF induced ε = -
Y
X
An induced current flows through the coil causing it to act
as an electromagnet. What pole will the end, X, of the coil
become, and why?
dΦ
dt = -B! v
v
(b) If the bar magnet is pulled away from the coil, what will
happen to the direction of the induced current, and why?
Answer:
(a) X will become a North pole (Lenz’s Law) to repel the
approaching magnet and slow down its approach. (If X
became a South pole, it would cause the magnet to
accelerate, increasing the magnitude of the induced current
and electrical energy in the circuit. There would be no added
mechanical work being done to produce this extra energy,
breaking the Law of Conservation of Energy.)
Example: Find the flux linkage when a 0.10T magnetic field is
directed at right angles through a rectangular coil (10cm by
6cm) with 25 turns on the coil.
B = 0.10T
6cm
10cm
(b) The induced current reverses direction. X becomes a South
pole, attracting the magnet and slowing it down as it moves
away. Work must be done to pull the magnet away against
this attraction - this work is the source of the electrical
energy produced in the circuit.
25 turns
Answer: Φ = nBA = 25 × 0.10 × (0.06 × 0.10) = 1.5 × 10-2Wb.
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Physics Factsheet
97. A2 Electromagnetic Problems
Search Coils
Example: Find the angular velocity of a coil rotating at a
frequency of 650 revolutions per minute (rpm).
When a search coil is pulled out of a magnetic field, the total charge
that flows into a ballistic galvanometer is measured. This can be
used to measure the strength of the magnetic field.
N
Answer: 650rpm = (650 × 2π) / 60 = 68rad s-1.
S
Example: Sketch the output voltage before and after the rate of
rotation of an a.c. generator is doubled.
Answer:
before
Ballistic Galvanometer
after
B = QR / An
where Q is the total charge measured, R is the total circuit
resistance, n is the number of turns in the search coil, and A
is the area of the coil (perpendicular to the field).
Remember that changing the rate of rotation of a
generator coil will affect both the frequency and amplitude of
the induced emf (and current). Changing either the strength of
the magnetic field or the number of turns on the coil will only
affect the amplitude of the induced emf.
Example: As a search coil is removed from a magnetic field,
the following graph is obtained.
distance/cm
Transformers
6.0
E.m. induction allows us to step-up or step-down an a.c. voltage
using a transformer. The basic relationships are:
4.0
Vs / Vp = Ns / Np = Ip / Is
2.0
where N stands for the number of turns on the coil.
t/ms
100
These relationships hold for an ideal transformer. This
implies no energy loss from the primary to the secondary coil.
For an ideal transformer: Pout = Pin
200
If a deflection of 1.0cm indicates a charge of 0.2µC, estimate
the total charge induced through the circuit.
Example: A mains transformer is used to reduce the output
voltage of a circuit to 12V (for use in an electronic device).
Answer:
The total charge is measured at the peak of the graph.
Maximum amplitude = 6.1cm
so Q = 6.1 × 0.20 = 1.22µC.
240 V
a.c.
Remember that the peak value of the galvanometer
deflection is proportional to the total charge that flows into
the ballistic galvanometer. Resist the temptation to find the
area under the graph. The time value gives an indication of
how quickly the coil was pulled from the field, but it does not
(in theory) affect the peak value of the graph.
Load
6000
turns
(a) How many turns are wound onto the secondary coil?
(b) If the load draws a current of 0.40A, find the current drawn
from the supply.
(c) What have we assumed?
Answer:
(a) Ns / Np = Vs / Vp, Ns = 6000 × 12 / 240 = 300 turns.
(b) Ip / Is = Vs / Vp, Ip = 0.40 × 12 / 240 = 0.020A
(c) An ideal transformer. No energy loss.
Generators
As the coil spins between the poles of the magnet in an a.c. generator,
the rate of change of the magnetic flux cutting the plane of the coil
changes sinusoidally, So the induced emf is also sinusoidal:
Example: Why does a transformer not work on direct current?
ε = BAnω
ω sinω
ωt
Answer:
E.m. induction requires a changing magnetic flux, dΦ / dt or
d(BA) / dt. As the area, A, is constant, then there must be a
changing magnetic field, B, linking the two coils. Direct current
would give a steady magnetic field. There would be no induced
emf.
Exam Hint: Angular velocity can be given in many different
forms (see example below). Remember that it must be converted
to radians per second (rad s -1 ) before being used in
calculations.
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Physics Factsheet
97. A2 Electromagnetic Problems
Answers
Questions
1. (a) The output coil would be in series with a set of resistors. The
current through the resistors would cause Joule heating.
The overall energy transfer is KE to thermal energy.
1. (a) In e.m. braking, the rotating axle is used to induce an emf
across an output coil. Explain what else would be required in
the output circuit, and what energy transfer occurs.
(b) Would this braking be more effective at high or low speeds?
(b) At high speeds, when dΦ
Φ/dt (and the induced current) is a
maximum.
2. A conducting bar 120cm long is pulled at a steady speed or
3.1ms-1 across a perpendicular uniform magnetic field of 1.40T.
(a) Find the flux swept out in 4.0s.
(b) Find the emf induced between the ends of the bar.
(c) What force would be required to keep the bar moving at
steady speed?
(d) What force would be required to keep the bar moving at
steady speed of the ends of the bar were linked through a
25 ohm resistor?
2. (a) ε = BA = 1.40 × (1.2 × 3.1 × 4.0) = 20.8Wb
Φ/dt = -5.2V
(b) ε = -dΦ
(c) None. No induced current means no energy is being
converted to heat.
(d) I = ε/R = 5.2 / 25 = 0.21A
Work required per second = Energy converted to heat per
second
Fs/t = Fv = I2R, F = 0.36N
3. How is the induction process different in an a.c. generator
compared to a d.c. generator?
4. The output of an a.c. generator is given by:
ε = BAnω
ω sinω
ωt
3. It isn’t. The direction of the output is switched twice per cycle
in a d.c. generator.
Explain why the output is a maximum when the rotating coil is
parallel to the magnetic field.
Φ/dt, or in this case dA/dt. Although the area of the coil
4. ε ∝ dΦ
perpendicular to the field is a minimum, the rate of change of the
area (dA/dt) perpendicular to the field is a maximum at this instant.
B
N
5. (a) Pin is greater than Pout (25.2W, 18.0W)
This is a non-ideal transformer. Energy is being lost (to heat).
S
(b) Joule heating in the wires of the transformer.
Eddy currents induced in the iron core (laminations should
minimise this problem).
Or hysteresis losses in the core due to the changing
magnetic field.
5. The transformer shown gives the following readings.
a.c.
input
Load
Vp
Vs
Ip
Is
12V
45V
2.10A
0.40A
(a) Explain these results. (Calculate input and output power.)
(b) Name 2 ways in which energy can be lost in a transformer.
Acknowledgements:
This Physics Factsheet was researched and written by Paul Freeman
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
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ISSN 1351-5136
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