JEE Mains · Physics · STD 12 -6. Electromagnetic induction
Magnetic flux (in weber) in a closed circuit of resistance \(20\,\Omega\) varies with time \(t(s)\) as \(\phi=8 t ^{2}-9 t +5\). The magnitude of the induced current at \(t =0.25\,s\) will be \(...mA\)
- A \(249\)
- B \(248\)
- C \(247\)
- D \(250\)
Answer & Solution
Correct Answer
(D) \(250\)
Step-by-step Solution
Detailed explanation
\(\phi=8 t ^{2}-9 t +5\) \(emf\) \(=-\frac{ d \phi}{ dt }=-(16 t -9)\) At \(t =0.25 s\) \(Emf\) \(=-[(16 \times 0.25)-9]=5 V\) \(Current\) \(=\frac{\text { Emf }}{\operatorname{Re} s i s \operatorname{tance}}=\frac{5 V }{20 \Omega}\) \(=\frac{1}{4} A =\frac{1000}{4} mA =250 mA\)
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