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III. Long Answer Questions · Q11

Q.Show that the mutual inductance between a pair of coils is same (M12=M21M_{12} = M_{21}).

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Step 1. For two long co-axial solenoids (turn densities n1n_1, n2n_2; length l; smaller area A2A_2), current i1i_1 in solenoid 1 produces field B1=μ0n1i1B_1=\mu_0n_1i_1, threading solenoid 2's smaller area A2A_2.

Step 2. The flux linkage of solenoid 2 is N2Φ21=(n2l)(μ0n1i1A2)=μ0n1n2A2l i1N_2\Phi_{21}=(n_2l)(\mu_0n_1i_1A_2)=\mu_0n_1n_2A_2l\,i_1, giving M21=μ0n1n2A2lM_{21}=\mu_0n_1n_2A_2l.

Step 3. Working the reverse direction, current i2i_2 in solenoid 2 produces field B2=μ0n2i2B_2=\mu_0n_2i_2, which (since B2B_2 is negligible outside solenoid 2) links solenoid 1 only over the SAME smaller area A2A_2.

Step 4. The flux linkage of solenoid 1 is N1Φ12=(n1l)(μ0n2i2A2)=μ0n1n2A2l i2N_1\Phi_{12}=(n_1l)(\mu_0n_2i_2A_2)=\mu_0n_1n_2A_2l\,i_2, giving M12=μ0n1n2A2lM_{12}=\mu_0n_1n_2A_2l. …

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