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Lesson: Chapter 13 — Electromagnetism and electromagnetic induction

Putting the chapter together 2 of 3

The exercise

Worked answers to exercise 13.5.

Question 1

A step-down transformer

  1. The primary has 1000 turns and the secondary 100. 230 V AC is supplied to the primary; assume no energy is lost. Find the greatest potential difference that can be taken from the secondary.
  2. If a 5 A alternating current is supplied to the primary and the efficiency is 100%, find the current from the secondary.

The answer

  1. VS = 230 V × 100 / 1000 = 23 V
  2. IS = VP IP / VS = 230 V × 5 A / 23 V = 50 A

Question 2

Another step-down transformer

  1. The primary has 5000 turns and the secondary 500, with 230 V on the primary and 100% efficiency. Find the secondary potential difference.
  2. If the secondary gives 10 A, find the current supplied to the primary.

The answer

  1. VS = 230 V × 500 / 5000 = 23 V
  2. IP = VS IS / VP = 23 V × 10 A / 230 V = 1 A

Question 3

A 1 : 10 transformer

  1. The turns on the primary and secondary are in the ratio 1 : 10. 6 V AC is supplied to the primary and 20 A is to be taken from the secondary; efficiency 100%. Find the secondary potential difference.
  2. Find the current supplied to the primary.
  3. Find the ratio of the primary voltage to the secondary voltage.
  4. Find the ratio of the primary current to the secondary current.

The answer

  1. VS = 6 V × 10 = 60 V
  2. IP = 60 V × 20 A / 6 V = 200 A
  3. VP : VS = 6 : 60 = 1 : 10
  4. IP : IS = 200 : 20 = 10 : 1 — the current ratio is the voltage ratio turned upside down.

Question 4

A magnet and a solenoid

  1. Briefly define electromagnetic induction.
  2. The N pole of a bar magnet is brought quickly towards the coil and the centre-zero galvanometer G deflects to the right. Does the current flow from A to B or from B to A?
  3. Which way does the galvanometer deflect as the N pole is taken away from the solenoid?
  4. Which way does it deflect if the S pole is brought towards the solenoid?
  5. Write three factors on which the strength of the current through the galvanometer depends.

The answer

  1. An emf is induced in a conductor when the magnetic field through it changes — by moving a magnet near it, or the conductor in a field.
  2. It depends on which terminal makes this galvanometer swing right. Taking the usual convention that a centre-zero meter swings towards the terminal the current enters, and A as that terminal as drawn, the current flows from A to B. The answers below follow from this one either way.
  3. To the left — the field through the coil now decreases instead of increasing.
  4. To the left — the field entering the coil is reversed.
  5. The number of turns on the coil, the strength of the magnet, and the speed at which it is moved.

Question 5

The bicycle dynamo

  1. Name parts A, B, C and D.
  2. What principle does the dynamo work on?
  3. Explain how a bicycle dynamo works.
  4. Is its current direct or alternating?
  5. Sketch a graph of how its emf changes with time.
  6. Explain why the lamp's brightness changes with the speed of pedalling.
  7. Write the energy change when a dynamo lights a bicycle lamp.

The answer

  1. A — the knurled driving wheel; B — the magnet; C — the coil; D — the soft iron core.
  2. Electromagnetic induction.
  3. The tyre spins the wheel and the magnet; the changing field through the coil on the soft iron induces an emf and a current in the lamp circuit.
  4. Alternating: the field through the coil reverses as the magnet turns.
  5. A sine-shaped graph swinging positive and negative, as below.
  6. Faster pedalling spins the magnet faster, so the field changes faster, the emf and current are larger, and the lamp is brighter.
  7. Mechanical (kinetic) energy → electrical energy → light (and heat).
t (cycles) 0.25 0.75 1.25 1.75
V 1 -1 1 -1

Question 6

The moving-coil microphone

  1. Name A, B, C and D and explain what each part does.

The answer

  1. A — the diaphragm, which vibrates when sound strikes it; B — the coil, which vibrates with it in the magnet's field and has an emf induced in it; C — the magnet, which gives the field; D — the output wires, which carry the small alternating current to the amplifier.