A coil having 'n' turns and resistance 'R' Ω is connected with a galvanometer of resistance '4R' Ω. What is the induced current in the circuit when this combination (coil+galvanometer) is moved in time t from a magnetic flux \( \phi_1 \) to \( \phi_2 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{-n(\phi_2 - \phi_1)}{5Rt} \)
By Faraday's law, the emf induced in a coil of \( n \) turns is \( n \) times the rate of change of flux through one turn; the negative sign is Lenz's law.
\( \displaystyle \varepsilon = -n\,\frac{\phi_2 - \phi_1}{t} \)
The coil and the galvanometer are in series, so the total resistance of the circuit is
\( \displaystyle R_{\mathrm{total}} = R + 4R = 5R \)
\( \displaystyle I = \frac{\varepsilon}{R_{\mathrm{total}}} = \frac{-n(\phi_2 - \phi_1)}{5Rt} \)