Q 12-06-129JEE MainJEE Main 2019 (10 Jan, Shift 2)Easy
The self induced emf of a coil is $25$ volts. When the current in it is changed at uniform rate from $10\ \text{A}$ to $25\ \text{A}$ in $1\ \text{s}$, the change in the energy of the inductance is:
Answer: (C) $437.5\ \text{J}$
$\varepsilon = L\dfrac{dI}{dt}$: $25 = L\times\dfrac{15}{1}$, so $L = \dfrac53\ \text{H}$.
$$\Delta U = \tfrac12L\left(I_2^2 - I_1^2\right) = \tfrac12\cdot\tfrac53\,(625 - 100) = 437.5\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics