A coil of 100 turns and area \(0.01\ \mathrm{m^{2}}\) has a resistance of \(2\ \Omega\). It is connected to an external resistor of \(8\ \Omega\). A uniform magnetic field normal to the plane of the coil increases at a steady rate of \(0.5\ \mathrm{T\,s^{-1}}\). The potential difference across the \(8\ \Omega\) resistor is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.4 V
The rising field changes the flux through each turn, so an emf is induced in the coil. This emf drives a current through the coil and the resistor, which are in series.
\( N=100 \)
\( A=0.01\ \mathrm{m^{2}} \)
\( \frac{dB}{dt}=0.5\ \mathrm{T\,s^{-1}} \)
\( r=2\ \Omega \)
\( R=8\ \Omega \)
The emf induced in the coil is
\( \begin{aligned} \varepsilon &= NA\frac{dB}{dt} \\ &= 100\times0.01\times0.5 \\ &= 0.5\ \mathrm{V} \end{aligned} \)
The current in the circuit is
\( \begin{aligned} I &= \frac{\varepsilon}{R+r} \\ &= \frac{0.5}{8+2} \\ &= 0.05\ \mathrm{A} \end{aligned} \)
The potential difference across the resistor is
\( \begin{aligned} V &= IR \\ &= 0.05\times8 \\ &= 0.4\ \mathrm{V} \end{aligned} \)
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)
Summary – “… If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”