Solution:
The magnetic field at the center of the outer coil of radius \(R\) carrying current \(I\) is \(B = \frac{\mu_0 I}{2R}\). The flux through the inner coil of radius \(r\) is \(\Phi = B \cdot (\pi r^2) = \frac{\mu_0 \pi r^2 I}{2R}\). Thus, \(M = \frac{\Phi}{I} = \frac{\mu_0 \pi r^2}{2R}\).
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