A square coil of side 10 cm, 500 turns and resistance \( 3\,\Omega \) is placed with its plane perpendicular to the horizontal component of the Earth's magnetic field. It is rotated about vertical axis passing through centre by \( 180^\circ \) in 0.25 second. Estimate the magnitude of the emf and current induced in the coil. Horizontal component of the Earth's magnetic field at the place is \( 3.0\times10^{-5}\,\mathrm{T} \).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 12\times10^{-4}\,\mathrm{V} \), \( 4\times10^{-4}\,\mathrm{A} \)
By Faraday's law the average emf is the change in flux linkage divided by the time. With its plane perpendicular to the field, the coil's flux is \( NBA \); after a turn of \( 180^\circ \) the field enters from the other face, so the flux becomes \( -NBA \).
\( \displaystyle A = (10\,\mathrm{cm})^{2} = (0.1\,\mathrm{m})^{2} = 1\times10^{-2}\,\mathrm{m^2} \)
\( \displaystyle \Delta\Phi = NBA - (-NBA) = 2NBA = 2 \times 500 \times 3.0\times10^{-5} \times 10^{-2} = 3\times10^{-4}\,\mathrm{Wb} \)
\( \displaystyle \varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{3\times10^{-4}}{0.25} = 1.2\times10^{-3}\,\mathrm{V} = 12\times10^{-4}\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{12\times10^{-4}}{3} = 4\times10^{-4}\,\mathrm{A} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Faraday’s Law of Induction – …
