JEE Mains · Physics · STD 12 -6. Electromagnetic induction
In the given figure the magnetic flux through the loop increases according to the relation \(\phi_{B}(t)=10 t^{2}+20 t\), where \(\phi_{B}\) is in milliwebers and \(t\) is in seconds. The magnitude of current through \({R}=2 \,\Omega\) resistor at \({t}=5\, {s}\) is \(....\,{mA}\).

- A \(10\)
- B \(60\)
- C \(180\)
- D \(120\)
Answer & Solution
Correct Answer
(B) \(60\)
Step-by-step Solution
Detailed explanation
As we know induce emf \(|\epsilon|=\frac{{d} \phi}{{dt}}=20 {t}+20\, {mV}\) \(|{i}|=\frac{|\epsilon|}{{R}}=10 {t}+10\, {mA}\) \(\text { at } {t}=5\) \(\mid {i} \mid=60\, {mA}\)
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