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GCSE / IGCSE Physics – Magnetism and Electromagnetism

Magnetic fields, motors, induction, transformers.

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12 questions  ·  38 marks  ·  answers included  ·  GCSE/IGCSE (Edexcel/AQA/Cambridge)
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This worksheet covers:

  • Practice questions covering magnetic fields
  • The motor effect
  • Electromagnetic induction and transformers
  • Matching the GCSE/IGCSE Physics specification

A calculator is allowed throughout.

Questions

Section A – Magnetic Fields

1

Describe the shape of the magnetic field pattern around a long straight current-carrying wire, and state how the pattern changes if the direction of the current is reversed.

Show answer
The field forms concentric circles around the wire, centred on the wire, in a plane perpendicular to it. Reversing the current reverses the direction of the field (the circles point the opposite way around the wire).
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2

Describe how a plotting compass can be used to plot the magnetic field pattern around a bar magnet.

Show answer
Place the plotting compass at a point near the magnet and mark the direction the compass needle points; move the compass a small distance in that direction and mark the new position, repeating this to build up a line of dots showing a field line from the north pole to the south pole. Repeat starting from different points around the magnet to build up the full field pattern.
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3

State one similarity and one difference between the magnetic field pattern of a bar magnet and the magnetic field pattern of a solenoid carrying a direct current.

Show answer
Similarity: both produce a field pattern like that of a bar magnet outside the solenoid/magnet, with field lines going from north to south. Difference: inside the solenoid, the field lines are parallel and evenly spaced (a strong, uniform field), whereas a bar magnet has no equivalent uniform interior field region.
[2]

Section B – The Motor Effect

4

State the two factors, other than the strength of the magnetic field, that affect the size of the force on a current-carrying wire placed in a magnetic field.

Show answer
The size of the current in the wire, and the length of the wire in the magnetic field (the force also depends on the angle between the wire and the field, being maximum when the wire is perpendicular to the field).
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5

A straight wire of length $0.12\,\text{m}$ carries a current of $3.0\,\text{A}$. It is placed perpendicular to a magnetic field of flux density $0.25\,\text{T}$. Calculate the force on the wire.

Show answer
$F = BIL = 0.25 \times 3.0 \times 0.12 = 0.090\,\text{N}$.
[3]
6

Explain how a simple DC electric motor uses the motor effect to produce continuous rotation of a coil, including the role of the split-ring commutator.

Show answer
A current-carrying coil is placed in a magnetic field; the forces on the two sides of the coil (in opposite directions, found using Fleming’s left-hand rule) create a turning effect (torque) that rotates the coil. The split-ring commutator reverses the direction of the current in the coil every half turn, so the direction of the forces (and hence the turning effect) stays the same, allowing the coil to keep rotating continuously in the same direction rather than oscillating.
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7

A rectangular current-carrying coil is placed in a uniform magnetic field, with the plane of the coil parallel to the field. Explain, in terms of the forces on the two sides of the coil, why the coil experiences a turning effect rather than moving off in a straight line.

Show answer
The current flows in opposite directions along the two sides of the coil that are perpendicular to the field, so by Fleming’s left-hand rule the forces on these two sides are equal in size but opposite in direction. These two equal and opposite forces, acting on opposite sides of the coil, form a couple that produces a turning effect (torque) rather than a resultant force, causing the coil to rotate.
[3]

Section C – Induction and Transformers

Calculator required
8

State the factors that affect the size of an electromotive force (EMF) induced in a coil by a changing magnetic field, and give two ways of increasing the size of an induced EMF.

Show answer
The induced EMF depends on the rate of change of magnetic flux and the number of turns on the coil. It can be increased by moving the magnet/coil faster (increasing the rate of change of flux) or by using a coil with more turns (or a stronger magnet).
[3]
9

A coil with $200$ turns experiences a change in magnetic flux of $0.030\,\text{Wb}$ over a time of $0.50\,\text{s}$. Calculate the magnitude of the EMF induced in the coil.

Show answer
$\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 200 \times \dfrac{0.030}{0.50} = 12\,\text{V}$.
[3]
10

A transformer has $1000$ turns on its primary coil and $50$ turns on its secondary coil. The primary coil is connected to a $230\,\text{V}$ a.c. supply.(a) Calculate the output (secondary) voltage.(b) If the current in the secondary coil is $4.0\,\text{A}$, calculate the current in the primary coil, assuming the transformer is $100\%$ efficient.

Show answer
(a) $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \Rightarrow V_s = 230 \times \dfrac{50}{1000} = 11.5\,\text{V}$.(b) For $100\%$ efficiency, $V_pI_p = V_sI_s \Rightarrow I_p = \dfrac{11.5 \times 4.0}{230} = 0.20\,\text{A}$.
[5]
11

Explain why transformers work with alternating current (a.c.) but do not work with direct current (d.c.).

Show answer
A transformer relies on electromagnetic induction: the changing current in the primary coil produces a changing magnetic flux in the iron core, which induces an EMF in the secondary coil. With d.c., the current (and therefore the magnetic flux) is constant and does not change, so no EMF is induced in the secondary coil.
[3]
12

Explain, in terms of energy losses, why electrical energy is transmitted through the National Grid at high voltage and low current, using step-up and step-down transformers.

Show answer
Power loss in the transmission cables is given by $P = I^2R$, so transmitting at a low current (with a correspondingly high voltage, since power transmitted $= IV$ stays the same) greatly reduces the energy wasted as heat in the cables due to their resistance. Step-up transformers increase the voltage (and reduce the current) for transmission, and step-down transformers reduce it again to a safe, usable voltage before it reaches homes and businesses.
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