Two concentric circular coils of radii 2 mm and 100 cm are placed co-axially. The mutual inductance of this arrangement will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 7.9 X 10−12 H
Pass a current I through the large coil (radius R). The field at its centre is \( \mu_0 I/2R \), and because the small coil (radius r) is very much smaller than R, this field is nearly uniform over the small coil. The flux through the small coil divided by I is the mutual inductance (single-turn coils).
\( \displaystyle B = \frac{\mu_0 I}{2R} \)
\( \displaystyle \Phi = B\pi r^{2} = \frac{\mu_0 I \pi r^{2}}{2R} \)
\( \displaystyle M = \frac{\Phi}{I} = \frac{\mu_0 \pi r^{2}}{2R} \)
\( r = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( R = 100\,\mathrm{cm} = 1\,\mathrm{m} \)
\( \displaystyle M = \frac{(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}) \times \pi \times (2\times10^{-3}\,\mathrm{m})^{2}}{2 \times 1\,\mathrm{m}} \)
\( \displaystyle M = 8\pi^{2}\times10^{-13}\,\mathrm{H} \approx 79\times10^{-13}\,\mathrm{H} \)
\( \displaystyle M \approx 7.9\times10^{-12}\,\mathrm{H} \)
Note: NTA picked this question from an NCERT (2026-27) example and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Mutual Inductance – …
