Solution:
Let a current \(I\) flow through the large circular coil, producing a magnetic field \(B = \frac{mu_0 I}{2R}\) at its center. The flux through the small square coil is \(\Phi = B'A = \left(\frac{\mu_0 I}{2R}\right) l^2\). Therefore, the mutual inductance is \(M = \frac{\Phi}{I} = \frac{\mu_0 l^2}{2R}\).
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