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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2022
  • Medium
  • Numerical

A \(10\, \Omega, 20 \; mH\) coil carrying constant current is connected to a battery of \(20 \; V\) through a switch is opened current becomes zero in \(100 \; \mu s\). The average emf induced in the coil is \(\dots \; V\).

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Answer & Solution

Correct Answer

(D) \(400\)

Step-by-step Solution

Detailed explanation

\(<\varepsilon>=\frac{\int \varepsilon dt }{\int dt }=\frac{\int( Ldi / dt ) dt }{\int dt }=\frac{ L \int di }{\int dt }\)

\(<\varepsilon>=\frac{ L \Delta i }{\Delta i }\)

\(i _{0}=\frac{ V }{ R }=\frac{20}{10}=2 \; A , \text { if } i =0 \; A\)

\(T =100 \; \mu s , L =20 \; mH\)

\(< \varepsilon>=\frac{20 \times 10^{-3} \times(2-0)}{100 \times 10^{-6}}.\)

\(=\frac{2 \times 10^{3}}{5}\)

\(< \varepsilon > =400 \; V\)

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