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.
Answer (integer)
250
Solution
<p>$R = 20\,\Omega$</p>
<p>$\phi = 8{t^2} - 9t + 5$</p>
<p>$$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = |16t - 9| = |16(0.25) - 9| = 5$$</p>
<p>$$i = {\varepsilon \over R} = {5 \over {20}} = 0.25\,A = {{0.25} \over {{{10}^3}}} \times {10^3}\,A = 250\,mA$$</p>
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Faraday's Laws
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