Case: Read the passage and answer the question.
Consider two long co-axial solenoids, each of length l. We denote the radius of the inner solenoid S1 by \( r_1 \) and its number of turns per unit length by \( n_1 \). The corresponding quantities for the outer solenoid S2 are \( r_2 \) and \( n_2 \), respectively. Let \( N_1 \) and \( N_2 \) be the total numbers of turns of coils S1 and S2, respectively.
A current \( I_1 \) is now passed through the inner solenoid S1. To find the flux linked with each turn of the outer solenoid S2, the area to be used is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \pi r_1^{2} \), the cross-section of S1
The flux through a turn is the field times the area over which that field exists. The solenoids are very long, so the field of S1 stays inside S1.
Option (1) \( \pi r_2^{2} \), the full cross-section of S2 — Incorrect. The field of S1 does not fill the whole of S2. Between the two solenoids it is almost zero.
Option (2) \( \pi r_1^{2} \), the cross-section of S1 — Correct. The field \( \mu_0n_1I_1 \) of S1 exists only over its own cross-section. So each turn of S2 links the flux \( \mu_0n_1I_1\,\pi r_1^{2} \).
Option (3) \( \pi(r_2^{2}-r_1^{2}) \), the space between S1 and S2 — Incorrect. This ring is the region where the field of S1 is almost zero. It adds nothing to the flux.
Option (4) \( \pi(r_1^{2}+r_2^{2}) \), the sum of both cross-sections — Incorrect. Outside S1, the field of S1 is almost zero. Adding the area of S2 counts flux that is not there.
NCERT (2026-27):
Mutual inductance – “The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoids are very long. Thus, flux linkage with solenoid S2 is”
\( N_2\Phi_2=(n_2l) \)\( (\pi r_1^{2}) \)\( (\mu_0n_1I_1) \)