Thank you for reporting, we will resolve it shortly
Q.
By a change of current from $5 \,A$ to $10\, A$ in $0.1\,s$, the self induced emf is $10\, V$. The change in the energy of the magnetic field of a coil will be
Electromagnetic Induction
Solution:
$\left|\varepsilon\right|=L \frac{\Delta I}{\Delta t}$
$L=\frac{\left|\varepsilon\right|\Delta t}{\Delta I}=\frac{10\times0.1}{\left(10-5\right)}=0.2\,H$
The magnetic field energies for currents $I_{1}$ and $I_{2}$ are
$U_{1}=\frac{1}{2}LI_{1}^{2}$ and $U_{2}=\frac{1}{2}LI_{2}^{2}$
Change in energy $= U_{2}- U_{1}$
$=\frac{1}{2}LI_{2}^{2}-\frac{1}{2}LI_{2}^{2}=\frac{L}{2}\left(I_{2}^{2}-I_{1}^{2}\right)$
$=\frac{0.2}{2}\left(10^{2}-5^{2}\right)$
$=7.5\,J$