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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2019
  • Hard
  • MCQ

The self produced \(emf\) of a coil is \(25\,volts\). When the current in it is changed at uniform rate from \(10\,A\) to \(25\, A\) in \(1\,s\), the change in the energy of the inductance is......\(J\)

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Answer & Solution

Correct Answer

(A) \(437.5\)

Step-by-step Solution

Detailed explanation

\(\frac{L \Delta T}{\Delta t}=25\)

\(\Rightarrow  L=\frac{25 \times 1}{15}=\frac{5}{3}\)

\(\Delta U=\frac{1}{2} L\left(l_{f}^{2}-l_{1}^{2}\right)=\frac{1}{2} \times \frac{5}{3} \times\left(25^{2}-10^{2}\right)\)

\(\quad=\frac{5}{6} \times 525=437 .5\, \mathrm{J}\)

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