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......\(J\)
Pick an option to check your answer.
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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