A \(10\ \mu\mathrm{F}\) capacitor is connected in series in a wire loop of area 100 cm2. The loop is held normal to a uniform magnetic field. The field increases steadily at \(0.2\ \mathrm{T\,s^{-1}}\). The charge on the capacitor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\times10^{-8}\ \mathrm{C}\)
The steadily changing field induces a constant emf in the loop. A steady current cannot flow through the capacitor. So the capacitor charges until the voltage across it equals the induced emf. Then its charge is \( q=C\varepsilon \).
\( \begin{aligned} A &= 100\ \mathrm{cm^{2}} = 10^{-2}\ \mathrm{m^{2}} \\ \frac{dB}{dt} &= 0.2\ \mathrm{T\,s^{-1}} \\ C &= 10\ \mu\mathrm{F} = 10^{-5}\ \mathrm{F} \end{aligned} \)
\( \begin{aligned} \varepsilon &= A\frac{dB}{dt} \\ &= 10^{-2}\times0.2 \\ &= 2\times10^{-3}\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} q &= C\varepsilon \\ &= 10^{-5}\times2\times10^{-3} \\ &= 2\times10^{-8}\ \mathrm{C} \end{aligned} \)
NCERT (2026-27):
Faraday’s Law of Induction – “Mathematically, the induced emf is given by”
\( \varepsilon=-\displaystyle\frac{d\Phi_B}{dt} \)\( \qquad (6.3) \)