 Solution

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

For chosen positive terminal currents, define self-inductance by the coil's own flux linkage

Λ_{j} = L_{j} I_{j}

Λ_{2} = M I_{1}

when only coil A carries current. With matching winding directions and terminal references,

L_{A} = \frac{μ _{0} n _{1}^{2} A _{t}}{s}, L_{B} = \frac{μ _{0} n _{2}^{2} A _{t}}{s}, M = \frac{μ _{0} n _{1} n _{2} A _{t}}{s} .

The reciprocal flux-linkage relation gives the same

M

from B to A, and

M^{2} = L_{A} L_{B}

in this ideal fully coupled long-solenoid limit. The induced emf is

- d Λ/ d t

; passive terminal voltage drops use the opposite sign,

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

and similarly for B.

 New to topics? Read the docs here!