Question

A toroidal solenoid with air core has an average radius \(R\), number of turns \(N\) and area of cross-section \(A\). The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid)

• A. \( \dfrac{\mu_0 N^2 A}{R} \)
• B. \( \dfrac{\mu_0 N^2 A}{2\pi R} \)
• C. \( \dfrac{\mu_0 N A}{2\pi R} \)
• D. \( \dfrac{\mu_0 N A}{R} \)

Hint:

For toroidal solenoids, always remember that the magnetic field depends on the mean radius \(R\).

Answer 1

The Correct Option is B

Step 1: Magnetic field inside a toroidal solenoid.
For a toroid with air core, the magnetic field at mean radius \(R\) is given by \[ B = \frac{\mu_0 N I}{2\pi R} \]

Step 2: Magnetic flux through one turn.
\[ \phi = B \cdot A = \frac{\mu_0 N I A}{2\pi R} \]

Step 3: Calculate total flux linkage.
\[ \Phi = N\phi = \frac{\mu_0 N^2 I A}{2\pi R} \]

Step 4: Use definition of self-inductance.
\[ L = \frac{\Phi}{I} = \frac{\mu_0 N^2 A}{2\pi R} \]