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

We are told that the self-induced emf of the coil is $$e = 25\ \text{V}$$. By definition of self-induction, the induced emf in an inductor of inductance $$L$$ is related to the time rate of change of current by the formula $$e = L\dfrac{dI}{dt}$$. First we determine $$L$$ from this relation.

The current rises uniformly from $$I_1 = 10\ \text{A}$$ to $$I_2 = 25\ \text{A}$$ in a time interval $$\Delta t = 1\ \text{s}$$, so the rate of change of current is

$$\dfrac{dI}{dt} = \dfrac{I_2 - I_1}{\Delta t} = \dfrac{25\ \text{A} - 10\ \text{A}}{1\ \text{s}} = 15\ \text{A s}^{-1}.$$

Substituting $$e = 25\ \text{V}$$ and $$\dfrac{dI}{dt}=15\ \text{A s}^{-1}$$ into $$e = L\dfrac{dI}{dt}$$, we obtain

$$L = \dfrac{e}{\dfrac{dI}{dt}} = \dfrac{25\ \text{V}}{15\ \text{A s}^{-1}} = \dfrac{5}{3}\ \text{H} = 1.666\dots\ \text{H}.$$

Now the magnetic energy stored in an inductor carrying current $$I$$ is given by the formula $$U = \dfrac12 L I^2$$. Therefore, the change in energy when the current changes from $$I_1$$ to $$I_2$$ is

$$\Delta U = \dfrac12 L\left(I_2^2 - I_1^2\right).$$

Substituting $$L = \dfrac{5}{3}\ \text{H}$$, $$I_2 = 25\ \text{A}$$ and $$I_1 = 10\ \text{A}$$, we get

$$\Delta U = \dfrac12 \left(\dfrac{5}{3}\right)\Bigl(25^2 - 10^2\Bigr) = \dfrac12 \left(\dfrac{5}{3}\right)(625 - 100) = \left(\dfrac{5}{6}\right)(525).$$

Simplifying the arithmetic,

$$\left(\dfrac{5}{6}\right)(525) = \dfrac{5 \times 525}{6} = \dfrac{2625}{6} = 437.5\ \text{J}.$$

Hence, the correct answer is Option C.