A long solenoid has 2000 turns per metre. Its cross-sectional area is 10 cm2. A coil of 10 turns and area 50 cm2 is wound over the middle of the solenoid, on the same axis. The current in the solenoid changes at the rate of \(25\ \mathrm{A\,s^{-1}}\). The magnitude of the emf induced in the coil is (Take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(6.3\times10^{-4}\ \mathrm{V}\)
The field of a long solenoid is \( B=\mu_0 nI \), and it is present only inside the solenoid. So the flux through each turn of the outer coil passes only through the cross-section of the solenoid. The part of the coil outside the solenoid has no field through it.
\( \begin{aligned} N &= 10 \\ n &= 2000\ \mathrm{m^{-1}} \\ A &= 10\ \mathrm{cm^{2}} = 10^{-3}\ \mathrm{m^{2}} \\ \frac{dI}{dt} &= 25\ \mathrm{A\,s^{-1}} \end{aligned} \)
The flux through one turn is \( \Phi_B=\mu_0 nIA \). By Faraday’s law,
\( \begin{aligned} \varepsilon &= N\frac{d\Phi_B}{dt} \\ &= N\mu_0 nA\frac{dI}{dt} \\ &= 10\times4\pi\times10^{-7}\times2000\times10^{-3}\times25 \\ &= 6.3\times10^{-4}\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Faraday’s Law of Induction – “In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, the expression for the total induced emf is given by”
\( \varepsilon=-N\displaystyle\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)
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. …”