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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/9/v/solution

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Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 9 / v / Solution by

Codex  0 2026-10-25

Use north-minus-south terminal voltages for both oppositely wound coils, with coil A measured at high input impedance so its AC current is zero. Then

v_{B} = L_{B} \dot{I}_{B}

and

v_{A} = M \dot{I}_{B}

, with

M < 0

. Hence

v_{A} (t) = \frac{M}{L _{B}} v_{B} (t) = - \frac{n _{1}}{n _{2}} V_{0} sin ω t .

(1)

The magnitude is the turns ratio times the drive, and these terminal references give a phase reversal. Reversing one voltmeter lead reverses the displayed sign. A fixed DC bias in A does not change this incremental result, while short-circuiting A would require solving both coupled current equations.

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