Mutual Inductance of Two Co-axial Coils – Rankers Physics

Mutual Induction: Practice Problem & Solution

Two circular coils one of small radius \(r\) and other of larger radius \(R\) (\(r \ll R\)) are placed co-axially with centres coinciding. The mutual inductance of the arrangement is
\(\frac{\mu_0 \pi R^2}{r}\)
\(\frac{\mu_0 \pi r^2}{2R}\)
\(\frac{\mu_0 \pi r R}{r + R}\)
\(\frac{2 \mu_0 \pi r^2}{R}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Mutual Induction. Understanding the underlying formula is key to arriving at the correct answer below:

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, \(M = \frac{\Phi}{I} = \frac{\mu_0 \pi r^2}{2R}\).

Leave a Reply

Your email address will not be published. Required fields are marked *