A long solenoid with 25 turns per cm has a small loop of area \( 2.0\,\mathrm{cm^2} \) placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 3.0 A to 5.0 A in 0.1 s. The induced emf in the loop due to change in the current is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\pi \times 10^{-6}\,\mathrm{V} \)
Inside a long solenoid the field is \( B = \mu_0 n I \). The changing current changes this field, so the flux through the loop changes, and Faraday's law gives the emf.
\( n = 25\,\mathrm{cm^{-1}} = 2500\,\mathrm{m^{-1}} \)
\( A = 2.0\,\mathrm{cm^2} = 2.0\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle \frac{dI}{dt} = \frac{5.0 - 3.0}{0.1} = 20\,\mathrm{A\,s^{-1}} \)
\( \displaystyle \frac{dB}{dt} = \mu_0\,n\,\frac{dI}{dt} = 4\pi\times10^{-7} \times 2500 \times 20 = 2\pi\times10^{-2}\,\mathrm{T\,s^{-1}} \)
\( \displaystyle \varepsilon = A\,\frac{dB}{dt} = 2.0\times10^{-4} \times 2\pi\times10^{-2} = 4\pi\times10^{-6}\,\mathrm{V} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
