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A-Level Magnetic flux, the induction laws of Faraday and Lenz, and power transformers

What the A-Level syllabus expects for Magnetic flux, the induction laws of Faraday and Lenz, and power transformers, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, identify, state. About 6% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (2 marks)

State Faraday's law of electromagnetic induction.

Show the worked answer

Faraday's law of electromagnetic induction: the magnitude of the induced e.m.f. is directly proportional to (equal to) the rate of change of magnetic flux linkage through the circuit.

Example 2 (2 marks)

Consider the four level rotor blades turning within the Earth's magnetic field. Give the potential difference between the far tips of two blades that point in exactly opposite directions, and justify your answer.

Show the worked answer

Consider two diametrically opposite blades of the same rotor turning in the Earth's field. Each blade acts as a rod sweeping through the field, so a motional emf is induced between the hub and each tip. For the two opposite blades, both the velocity v and the radial direction (dl from hub to tip) reverse together, so the integral of (v x B).dl from hub to tip has the same sign and magnitude for each blade. Hence both tips reach the same potential relative to the hub. The potential difference between the two far tips is therefore the difference of two equal potentials = 0.

Example 3 (2 marks)

For an oscillating LC circuit (C = 2.0 μF charged initially to 40 V, L = 8.0 mH): at the moment the capacitor becomes completely discharged, the circuit current reaches its maximum value. Work out this maximum current in the circuit.

Show the worked answer

Energy is conserved in the LC circuit. When the capacitor is fully discharged all its initial electrical energy is stored in the inductor as magnetic energy, and the current is maximum. (1/2) C V² = (1/2) L I_max² I_max = V sqrt(C/L) = 40 x sqrt(2.0x10⁻⁶ / 8.0x10⁻³) = 40 x sqrt(2.5x10⁻⁴) = 40 x 0.01581 = 0.63 A.

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