Two coils are placed close to each other. Coil 1 has a self-inductance of 0.4 H, and the mutual inductance of the pair is 0.1 H. At an instant, the current in coil 1 is increasing at \(5\ \mathrm{A\,s^{-1}}\) and the current in coil 2 at \(10\ \mathrm{A\,s^{-1}}\). The fluxes of the two currents through coil 1 are in the same direction. The magnitude of the emf induced in coil 1 at this instant is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\ \mathrm{V}\)
When both coils carry currents, the flux through coil 1 has two parts. One part comes from its own current and the other from the current in coil 2. Here both parts grow in the same direction, so their emfs add.
\( L_1=0.4\ \mathrm{H} \)
\( M_{12}=0.1\ \mathrm{H} \)
\( \begin{aligned} |\varepsilon_1| &= L_1\frac{dI_1}{dt}+M_{12}\frac{dI_2}{dt} \\ &= 0.4\times5+0.1\times10 \\ &= 2+1 \\ &= 3\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. …”
“M11 is the self-inductance and is written as L1. Therefore,”
\( \displaystyle \varepsilon_1=-L_1\frac{dI_1}{dt}-M_{12}\frac{dI_2}{dt} \)