Magnetic flux (in weber) in a closed circuit of resistance 20 Ω varies with time t(s) at φ = 8t2 - 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA. Numerical answer type. Work it out, then open the answer.

Answer: 250. e = |(dφ)/(dt)| with φ = 8t² - 9t + 5, so (dφ)/(dt) = 16t - 9. At t = 0.25 s: (dφ)/(dt) = 4 - 9 = -5 V, so |e| = 5 V. I = e/R = 5/20 = 0.25 A So the induced current is 250 mA.

Magnetic flux (in weber) in a closed circuit of resistance 20 Ω… | JEE Main 2022 (25 Jul, Shift 2)

Q55JEE Main 2022 · 25 JUL · SHIFT 2Easy

Magnetic Flux & Faraday's Law

Magnetic flux (in weber) in a closed circuit of resistance 20 $\Omega$ varies with time t(s) at $\phi$ = 8t2 $-$ 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.

Numerical answer type. Work it out, then open the answer.

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Answer: 250

$e = \left|\frac{d\phi}{dt}\right|$ with $\phi = 8t^2 - 9t + 5$ , so $\frac{d\phi}{dt} = 16t - 9$ .
At $t = 0.25$ s: $\frac{d\phi}{dt} = 4 - 9 = -5$ V, so $|e| = 5$ V.
$I = \frac{e}{R} = \frac{5}{20} = 0.25$ A
So the induced current is $250$ mA.

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