Question

A long solenoid of radius \(R\) and length \(L\) has \(n\) turns per unit length. A circular loop of radius \(r(<R)\) is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.

Answer 1

Mutual inductance of solenoid and circular loop

Step 1: Magnetic field inside long solenoid
For a long solenoid carrying current \(I\), magnetic field inside is uniform and given by: \[ B = \mu_0 n I \] where:
• \(n\) = number of turns per unit length
• \(\mu_0\) = permeability of free space

Step 2: Flux through circular loop
A circular loop of radius \(r\) is placed inside the solenoid with axis aligned. So area of loop: \[ A = \pi r^2 \] Magnetic flux through the loop: \[ \Phi = B \cdot A \] Substitute \(B\): \[ \Phi = (\mu_0 n I)(\pi r^2) \] \[ \Phi = \mu_0 n I \pi r^2 \]

Step 3: Definition of mutual inductance
Mutual inductance is: \[ M = \frac{\Phi}{I} \] Substitute flux: \[ M = \frac{\mu_0 n I \pi r^2}{I} \] Cancel \(I\): \[ M = \mu_0 n \pi r^2 \]

Step 4: Final result
\[ \boxed{M = \mu_0 n \pi r^2} \] Physical interpretation:
• Mutual inductance depends on geometry of system
• Independent of current \(I\)
• Proportional to area of loop and turn density of solenoid