The mutual inductance of two coils is 0.25 H. The current in the first coil falls steadily from 4 A to zero in 0.1 s. The second coil forms a closed circuit of resistance \(5\ \Omega\). The current induced in the second coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\ \mathrm{A}\)
A changing current in one coil induces an emf \( \varepsilon=M\,dI/dt \) in a nearby coil. The second coil is closed, so this emf drives a current \( I=\varepsilon/R \) through it.
\( \begin{aligned} \frac{dI}{dt} &= \frac{4-0}{0.1} \\ &= 40\ \mathrm{A\,s^{-1}} \end{aligned} \)
\( \begin{aligned} \varepsilon &= M\frac{dI}{dt} \\ &= 0.25\times40 \\ &= 10\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} I &= \frac{\varepsilon}{R} \\ &= \frac{10}{5} \\ &= 2\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Mutual inductance – “… We get,”
\( \varepsilon_1=-M\displaystyle\frac{dI_2}{dt} \)
“It shows that varying current in a coil can induce emf in a neighbouring coil. …”
Summary – “… If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”