JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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......\(J\)
- A \(437.5\)
- B \(740\)
- C \(540\)
- D \(637.5\)
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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