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Q.

Magnetic flux (in weber) in a closed circuit of resistance $20 Ω$ varies with time t(s) as $ϕ = 8 t^{2} - 9 t + 5$ . The magnitude of the induced current at t = 0.25 s will be _________ mA.

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answer is 250.

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Detailed Solution

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$\begin{array}{l} ϕ = 8 t^{2} - 9 t + 5 \\ emf = - \frac{dϕ}{dt} = - (16 t - 9) \\ At t = 0 .25 s \\ Emf = - [(16 \times 0 .25) - 9] = 5 V \\ Current = \frac{Emf}{Re sis \tan ce} = \frac{5 V}{20 Ω} \\ = \frac{1}{4} A = \frac{1000}{4} mA = 250 mA \end{array}$

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Magnetic flux (in weber) in a closed circuit of resistance 20Ω varies with time t(s) as ϕ=8t2-9t+5. The magnitude of the induced current at t = 0.25 s will be _________ mA.