A coil having ' N ' turns and resistance ' R ' $\Omega$ is connected to a galvanometer of resistance '6R'…
- $\frac{\mathrm{N}\left(\phi_2-\phi_1\right)}{\mathrm{t}}$
- $\frac{\mathrm{N}\left(\phi_2-\phi_1\right)}{7 \mathrm{Rt}}$
- $\frac{\mathrm{N}\left(\phi_2-\phi_1\right)}{\mathrm{Rt}}$
- $\frac{N\left(\phi_2-\phi_1\right)}{6 R t}$
Solution
The induced current $I$ is determined by the induced EMF and the total circuit resistance.
With coil resistance $R$ and galvanometer resistance $6R$ in series, the total resistance is $R_{\text{total}} = R + 6R = 7R$.
Faraday's law gives the magnitude of induced EMF as $|\varepsilon| = N \frac{\phi_2 - \phi_1}{t}$ for flux change $\phi_2 - \phi_1$ over time $t$.
Applying Ohm's law yields $I = \frac{|\varepsilon|}{R_{\text{total}}} = \frac{N(\phi_2 - \phi_1)}{7Rt}$.
Final answer: $\boxed{\text{B}}$
Asked in: MHT CET 2025 (05 May Shift 2)
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