A short solenoid of radius a and \( n_1 \) number of turns per unit length and length L is kept coaxially inside a very long solenoid of radius b and \( n_2 \) number of turns per unit length. What is the mutual inductance of the system.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \mu_0\pi a^{2} n_1 n_2 L \)
The mutual inductance is the flux linked with one coil per unit current in the other, \( N\Phi = M I \). It is easiest to pass a current \( I \) through the long outer solenoid, whose field inside is uniform.
\( B = \mu_0\,n_2\,I \)
This field passes through each turn of the short inner solenoid over that turn's own area \( \pi a^{2} \). The inner solenoid has \( n_1 L \) turns.
\( \displaystyle N\Phi = (n_1 L)\,(\mu_0 n_2 I)\,(\pi a^{2}) \)
\( \displaystyle M = \frac{N\Phi}{I} = \mu_0\pi a^{2} n_1 n_2 L \)
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 – “… For these simple co-axial solenoids it is possible to calculate M12. The magnetic field due to the current I2 in S2 is μ0n2I2.”
Mutual Inductance – “… Thus, from Eq. (6.7) and Eq. (6.8),
\( \displaystyle M_{12} = \mu_0 n_1 n_2 \pi r_1^{2} l \qquad (6.9) \)”
Mutual Inductance – “… Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage N1Φ1 because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid.”