A coil of 'n' turns and resistance R Ω is connected in series with a resistance R/4. The combination is moved for time 't' second through flux Φ₁ to Φ₂. The induced current in the circuit is (a) (4n(Φ₁ - Φ₂))/(5RT) (b) 5n²(Φ₁ - Φ₂)4RT (c) (2n(Φ₁ - Φ₂))/(3RT) (d) (3n(Φ₁ - Φ₂))/(4RT)

Answer: (a). Total resistance R + R/4 = 5R/4 Emf e = (n(Φ₁ - Φ₂))/(t) I = e/5R/4 = (4n(Φ₁ - Φ₂))/(5Rt) So the answer is (a) (4n(Φ₁ - Φ₂))/(5Rt).

A coil of 'n' turns and resistance R Ω is connected in series with a… | MHT CET 2026 (15 Apr, Shift 1)

Q7MHT CET 2026 · 15 APR · SHIFT 1Easy

Magnetic Flux & Faraday's Law

A coil of 'n' turns and resistance $R\,\Omega$ is connected in series with a resistance $R/4$ . The combination is moved for time 't' second through flux $\Phi_1$ to $\Phi_2$ . The induced current in the circuit is

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Answer: (a)

Total resistance $R + \frac{R}{4} = \frac{5R}{4}$
Emf $e = \frac{n(\Phi_1 - \Phi_2)}{t}$
$I = \frac{e}{5R/4} = \frac{4n(\Phi_1 - \Phi_2)}{5Rt}$
So the answer is (a) $\frac{4n(\Phi_1 - \Phi_2)}{5Rt}$ .

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Magnetic Flux & Faraday's Law question from MHT CET 2026 (15 Apr, Shift 1)

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