A coil of 100 turns and area \(0.01\ \mathrm{m^{2}}\) has a resistance of 10 Ω. Its plane is perpendicular to a uniform magnetic field of 0.5 T. The coil is turned through \(180^\circ\) about one of its diameters. The charge that flows through the coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.1\ \mathrm{C}\)
The charge that flows depends only on the change in flux linkage and on the resistance. Flux linkage is the number of turns times the flux through one turn. The time taken for the turn does not matter. Turning the coil through \( 180^\circ \) reverses the flux through it, so the change per turn is \( 2BA \).
\( \begin{aligned} Q &= I\,\Delta t \\ &= \frac{\varepsilon}{R}\,\Delta t \\ &= \frac{N\,\Delta\Phi}{R} \end{aligned} \)
\( \begin{aligned} \Delta\Phi &= BA-(-BA) \\ &= 2BA \\ &= 2\times0.5\times0.01 \\ &= 0.01\ \mathrm{Wb} \end{aligned} \)
\( \begin{aligned} Q &= \frac{100\times0.01}{10} \\ &= 0.1\ \mathrm{C} \end{aligned} \)
NCERT (2026-27):
Example 6.3 – “Initial flux through the coil, ΦB (initial) = BA cos θ = 3.0 × 10–5 × (π ×10–2) × cos 0° = 3π × 10–7 Wb”
“Final flux after the rotation, ΦB (final) = 3.0 × 10–5 × (π ×10–2) × cos 180° = –3π × 10–7 Wb”
“Therefore, estimated value of the induced emf is,”
\( \displaystyle \varepsilon=N\frac{\Delta\Phi}{\Delta t} \)\( =500\times(6\pi\times10^{-7})/0.25=3.8\times10^{-3}\ \mathrm{V} \)
\( I=\varepsilon/R=1.9\times10^{-3}\ \mathrm{A} \)