What is the mutual inductance of two conducting concentric circular loops of radii \( R_{1} \) and \( R_{2}\ (R_{2} \lt R_{1}) \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.5\,\mu_{0}\,(\pi R_{2}^{2})\,R_{1}^{-1} \)
Mutual inductance is found from \( \Phi_{2} = MI_{1} \): pass a current \( I_{1} \) through the larger loop (radius \( R_{1} \)) and find the flux through the smaller loop (radius \( R_{2} \)). The field of the larger loop at its centre is known exactly, and over the small area of the inner loop it can be taken as uniform.
\( \displaystyle B_{1} = \frac{\mu_{0}I_{1}}{2R_{1}} \quad \text{(field at the centre of the larger loop)} \)
\( \displaystyle \Phi_{2} = B_{1}(\pi R_{2}^{2}) = \frac{\mu_{0}I_{1}}{2R_{1}}\,\pi R_{2}^{2} \)
\( \displaystyle M = \frac{\Phi_{2}}{I_{1}} = \frac{\mu_{0}\pi R_{2}^{2}}{2R_{1}} = 0.5\,\mu_{0}\,(\pi R_{2}^{2})\,R_{1}^{-1} \)
Note: The question states only \( R_{2} \lt R_{1} \). The expression \( \mu_{0}\pi R_{2}^{2}/2R_{1} \) treats the field of the larger loop as uniform over the whole area of the smaller loop, which holds only when \( R_{2} \) is much smaller than \( R_{1} \); it seems that NTA has assumed this. For loops of comparable radii the exact mutual inductance is different, but none of the options gives it, so Option (1) is the intended answer.