A coil of 100 turns and area \(20\ \mathrm{cm^{2}}\) has a resistance of 4 Ω. Its plane is perpendicular to a uniform magnetic field. At what rate must the field change to drive a current of 0.05 A through the coil?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1\ \mathrm{T\,s^{-1}}\)
The current \( I \) needs an emf \( \varepsilon=IR \). The area of the coil does not change, so the flux through each turn changes only because \( B \) changes. Faraday’s law then gives \( \varepsilon=NA\,dB/dt \).
\( N=100 \)
\( A=20\ \mathrm{cm^{2}}=2\times10^{-3}\ \mathrm{m^{2}} \)
\( R=4\ \Omega \)
\( I=0.05\ \mathrm{A} \)
\( \begin{aligned} \varepsilon &= IR \\ &= 0.05\times4 \\ &= 0.2\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} \frac{dB}{dt} &= \frac{\varepsilon}{NA} \\ &= \frac{0.2}{100\times2\times10^{-3}} \\ &= 1\ \mathrm{T\,s^{-1}} \end{aligned} \)
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)