Two conducting circular loops of radii a and b are placed in the same plane with their centre coinciding. If a >> b, the mutual inductance M between them will be directly proportional to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{b^{2}}{a} \)
The mutual inductance is the flux linked with one loop per unit current in the other, \( M = \Phi/I \). Send a current I through the large loop (radius a). Since \( a \gg b \), the field over the small loop is almost the field at the centre of the large loop, which is uniform over the small area.
\( \displaystyle B = \frac{\mu_0I}{2a} \)
\( \displaystyle \Phi = B\times\pi b^{2} = \frac{\mu_0\pi b^{2}I}{2a} \)
\( \displaystyle M = \frac{\Phi}{I} = \frac{\mu_0\pi b^{2}}{2a} \propto \frac{b^{2}}{a} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Mutual Inductance – …
