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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2021
  • Medium
  • Numerical

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 \(....\,{mA}\).

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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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