A coil of 200 turns has a small cut in its wire, so no current can flow through it. The magnetic flux through each turn of the coil changes steadily at the rate of \(5\ \mathrm{mWb\,s^{-1}}\). The potential difference between the two ends of the wire at the cut is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1\ \mathrm{V}\)
A changing flux induces an emf in a coil whether the circuit is closed or open. The cut only stops the current. Each turn has the same rate of change of flux, so the emfs of the 200 turns add up.
\( N=200 \)
\( \frac{d\Phi_B}{dt}=5\times10^{-3}\ \mathrm{Wb\,s^{-1}} \)
\( \begin{aligned} |\varepsilon| &= N\frac{d\Phi_B}{dt} \\ &= 200\times5\times10^{-3} \\ &= 1\ \mathrm{V} \end{aligned} \)
No current flows, so there is no potential drop across the resistance of the wire. Hence, the potential difference across the cut is equal to the emf.
NCERT (2026-27):
Lenz’s Law and Conservation of Energy – “What will happen if an open circuit is used in place of the closed loop in the above example? In this case too, an emf is induced across the open ends of the circuit. …”
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”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)