Question

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

• A. 5 J
• B. 6 J
• C. 7.5 J
• D. 9 J

Answer 1

Answer: (C)

7.5 J

Answer 2

$|\varepsilon| = L\frac{\Delta I}{\Delta t}$

$L = \frac{|\varepsilon|\Delta t}{\Delta I} = \frac{10 \times 0.1}{(10 - 5)} = 0.2H$

The magnetic field energies for currents $I_{1}$and $I_{2}$ are

$U_{1} = \frac{1}{2}LI_{1}^{2}$ ,$U_{2} = \frac{1}{2}LI_{2}^{2}$

Change in energy $= U_{2} - U_{1}$

$= \frac{1}{2}LI_{2}^{2} - \frac{1}{2}LI_{2}^{2} = \frac{L}{2}(I_{2}^{2} - I_{1}^{2})$

$= \frac{0.2}{2}(10^{2} - 5^{2}) = 7.5J$