The problems are on the electromagnetism worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electromagnetism links electricity and magnetism: a current produces a magnetic field, a current-carrying wire in a field feels a force F = B I L, and a changing magnetic field induces a voltage. These worked answers correct the 20 problems of the Physics 9-10 worksheet on electromagnetism, from attract or repel? to magnet in a copper tube: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit, with final results in bold. Direction answers name the rule used. Keep intermediate values unrounded and round only the last step, as examiners expect.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS2-5, GCSE P7
Problem 1 Attract or repel?
1. Attract (unlike poles).2. Repel (like poles).
3. Repel (like poles).
4. Attract: a magnet induces magnetism in iron, and the nearer pole of the nail becomes the opposite pole.
Problem 2 Field line facts
1. From the north pole to the south pole.2. At the poles, where the lines are closest together.
3. It turns to line up with the field, its north end pointing along the field line.
4. Tesla (T).
Problem 3 Force on a wire
1. F = B I L = 0.50 × 2.0 × 0.10 = 0.10 N.2. 0 N: the wire does not cut across the field.
3. Force is proportional to current: 0.20 N.
Problem 4 Unit conversions
1. 25 ÷ 100 = 0.25 m.2. 250 ÷ 1000 = 0.25 T.
3. 40 ÷ 1000 = 0.040 A.
4. T = 1 ÷ f = 1 ÷ 50 = 0.020 s = 20 ms.
Problem 5 Stronger electromagnet
1. Increase the current, use more turns, and use a soft iron core.2. Soft iron is magnetised easily and loses its magnetism when the current stops. Steel would stay magnetised.
3. It can be switched off (and its strength can be varied by changing the current).
Problem 6 Transformer output
1. Vs = Vp × Ns ÷ Np = 230 × 50 ÷ 1000 = 11.5 V.2. Step-down, because the secondary has fewer turns.
3. Ns = Np × Vs ÷ Vp = 1000 × 46 ÷ 230 = 200 turns.
Problem 7 Making more voltage
1. Turn the wheel faster, use a stronger magnet, or use a coil with more turns.2. The coil or magnet turns, so each half turn the wire cuts the field in the opposite direction and the voltage reverses.
3. Nothing: with no relative movement the field does not change.
Problem 8 Finding the current
1. I = F ÷ (B L) = 0.36 ÷ (0.30 × 0.20) = 0.36 ÷ 0.060 = 6.0 A.2. B = F ÷ (I L) = 0.36 ÷ (3.0 × 0.20) = 0.60 T.
Problem 9 Direction of the force
1. The field points right and the current goes into the page, so the force is downwards.2. Reversing the current reverses the force: upwards.
3. Reversing both leaves the force unchanged: downwards.
4. Fleming’s left-hand rule.
Problem 10 Phone charger transformer
1. Ns = 2300 × 9.0 ÷ 230 = 90 turns.2. P = V I = 9.0 × 1.5 = 13.5 W.
3. Ip = P ÷ Vp = 13.5 ÷ 230 = 0.059 A.
Problem 11 Reading a dynamo graph
1. The peak is 6.0 V.2. One full cycle takes 40 ms, so the period is 40 ms = 0.040 s. f = 1 ÷ 0.040 = 25 Hz.
3. The curve is at its peak a quarter of a period after the start: 6.0 V.
4. The peak becomes 12 V and the period halves to 20 ms, so the frequency doubles to 50 Hz.
Problem 12 Why alternating?
1. A battery gives a steady current and a steady field, so the field through the secondary coil does not change and no voltage is induced.2. The flick occurs when the switch is closed: the field grows from zero, which is a change that induces a brief voltage.
3. The iron core guides and strengthens the changing field so almost all of it passes through the secondary coil.
Problem 13 Electromagnet data
1. The number of clips increases roughly in proportion to the number of turns, about 0.45 clips per turn.2. 50 × 0.45 = 22.5, so about 22 to 23 clips.
3. The current and the core material (also the size of the paper clips and the battery).
Problem 14 How much can the force lift?
1. F = B I L = 0.080 × 3.5 × 0.12 = 0.034 N (0.0336 N).2. m = F ÷ g = 0.0336 ÷ 9.8 = 0.00343 kg = 3.4 g.
3. Doubling I doubles F to 0.0672 N: m = 0.0672 ÷ 9.8 = 0.00686 kg = 6.9 g.
Problem 15 Motor against generator
1. Motor: electrical to kinetic. Generator: kinetic to electrical.2. Both have a coil that can turn between the poles of a magnet.
3. The motor uses a split-ring commutator to reverse the current every half turn so the turning effect keeps the same direction. The generator uses slip rings so that the alternating voltage is delivered unchanged.
Problem 16 Transformer efficiency
1. Ns ÷ Np = Vs ÷ Vp = 240 ÷ 12 = 20.2. Is = P ÷ V = 60 ÷ 240 = 0.25 A.
3. Input power = 60 ÷ 0.92 = 65 W (65.2 W).
4. Ip = 65.2 ÷ 12 = 5.4 A.
5. About 5.2 W is lost as heat in the coils (resistance) and in the core (eddy currents).
Problem 17 Lamp on a transformer
1. Vs = 24 × 2000 ÷ 400 = 120 V.2. Step-up, since Ns is larger than Np.
3. Is = P ÷ V = 60 ÷ 120 = 0.50 A.
4. Vp Ip = 60 W, so Ip = 60 ÷ 24 = 2.5 A.
5. Ip ÷ Is = 2.5 ÷ 0.50 = 5.0 and Ns ÷ Np = 2000 ÷ 400 = 5.0, so they are equal.
Problem 18 Spot the errors
1. Power cannot increase. In an ideal transformer the output power equals the input power, and in a real one it is smaller because of heating losses.2. A magnet at rest causes no change of field, so no voltage is induced. Voltage appears only while the magnet or the field moves.
3. A wire parallel to the field cuts no field lines, so the force is zero. The force is greatest at right angles.
Problem 19 Turning effect of a coil
1. F = N B I L = 50 × 0.40 × 1.2 × 0.060 = 1.44 N.2. It points in the opposite direction, since the current runs the other way in the field.
3. Moment = F × distance = 1.44 × 0.040 = 0.058 N m (0.0576 N m).
4. Zero, because the two forces then act along the same line.
5. The commutator reverses the current every half turn so that each side is pushed the same way round and the coil keeps spinning instead of stopping.
Problem 20 Magnet in a copper tube
1. t = √(2h ÷ g) = √(2 × 0.50 ÷ 9.8) = √0.102 = 0.32 s.2. t = 0.50 ÷ 0.25 = 2.0 s.
3. The moving magnet changes the field through the copper, inducing circulating currents. By Lenz’s law these produce a field that opposes the motion, so there is an upward force and the magnet falls slowly.
4. ΔE = m g h = 0.040 × 9.8 × 0.50 = 0.20 J (0.196 J).
5. P = 0.196 ÷ 2.0 = 0.098 W, transferred as heat in the tube.
