In a coil, the current is changed from 12 A to 24 A in 1 s. The self-induced emf of the coil is 24 V. The change in energy in the coil is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 432 J
The self-induced emf is \( \varepsilon = L\,|\Delta I|/\Delta t \), which gives the self-inductance. The energy stored in the magnetic field of the coil is \( U = \tfrac{1}{2}LI^{2} \).
\( \displaystyle L = \frac{\varepsilon\,\Delta t}{\Delta I} = \frac{24\times1}{24 - 12} = 2\,\mathrm{H} \)
\( \displaystyle \Delta U = \tfrac{1}{2}L\left(I_2^{2} - I_1^{2}\right) = \tfrac{1}{2}\times2\times\left(24^{2} - 12^{2}\right) \)
\( \displaystyle \Delta U = 576 - 144 = 432\,\mathrm{J} \)