Question

Two planar concentric rings of metal wire having radii '\(r_1\)' and '\(r_2\)' (with \(r_1 > r_2\)) are placed in air. The current 'I' is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( \(μ_0\) = permeability of free space)

• A. \(\frac{μ_0π\,r_1^2}{2\,r_2}\)
• B. \(\frac{μ_0π\,r_2^2}{2\,r_1}\)
• C. \(\frac{μ_0π\,(r_1+r_2)^2}{2\,r_1}\)
• D. \(\frac{μ_0π\,(r_1-r_2)^2}{2\,r_2}\)

Hint:

A rotating charge is equivalent to a current q f, and a circular loop gives mu_0 I over 2R at the centre.

Answer 1

The Correct Option is B

Step 1: Use the period:
The time period of rotation is $T = \frac1f$, so $I = \frac qT = qf$.

Step 2: Apply the loop formula:
For a circular current loop, $B = \frac{\mu_0I}{2R}$. Substituting gives $\frac{\mu_0qf}{2R}$ (C).

Final Answer:
$\frac{\mu_0qf}{2R}$. \[ \boxed{\frac{\mu_0 qf}{2R}} \]