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.
Let \( n_1=3000 \) turns per metre and \( n_2=1000 \) turns per metre. The areas of cross-section of S1 and S2 are 5 cm2 and 20 cm2, respectively. A current of 2 A flows in S2 only. The magnetic flux through each turn of S1 is (take \( \mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.26\times10^{-6}\ \mathrm{Wb} \)
The current in the outer solenoid S2 sets up a uniform field \( B=\mu_0n_2I_2 \) inside it. The inner solenoid S1 lies in this field. So the flux through each turn of S1 is this field times the area of S1.
Given:
\( n_2=1000\ \mathrm{m^{-1}} \)
\( I_2=2\ \mathrm{A} \)
\( A_1=5\ \mathrm{cm^{2}}=5\times10^{-4}\ \mathrm{m^{2}} \)
\( \begin{aligned} B &= \mu_0n_2I_2 \\ &= 4\pi\times10^{-7}\times1000\times2 \\ &= 2.51\times10^{-3}\ \mathrm{T} \end{aligned} \)
\( \begin{aligned} \Phi_1 &= BA_1 \\ &= 2.51\times10^{-3}\times5\times10^{-4} \\ &= 1.26\times10^{-6}\ \mathrm{Wb} \end{aligned} \)
The value of \( n_1 \) and the area of S2 are not needed here.
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. The resulting flux linkage with coil S1 is,”
\( N_1\Phi_1=(n_1l) \)\( (\pi r_1^{2})(\mu_0n_2I_2) \)\( =\mu_0n_1n_2\pi r_1^{2}l\,I_2 \qquad (6.8) \)