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

Options

A. 5J
B. 6J
C. 7.5J
D. 9J

Answer

C

Explanation

`|epsi| = L (Delta I)/(Delta t), L = (|epsi| Delta t)/(Delta I) = (10 xx 0.1)/((10-5)) = 0.2 H`
The magnetic field energies for currents `I_1` and `I_2` are
`U_1 = 1/2 LI_1^2 `and `U_2 = 1/2 LI_2^2`
Change in energy ` = U_2 - U_1`
` = 1/2 LI_2^2 - 1/2 LI_1^2 = L/2 (I_2^2 - I_1^2) = (0.2)/(2) (10^2- 5^2) = 7.5 J`