An emf of 0.5 V is induced in a coil when the current in it changes at the rate of 10 A s−1. What steady current must flow in this coil so that it stores 0.4 J of magnetic energy?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4\ \mathrm{A}\)
The size of the self-induced emf in a coil is \( \varepsilon=L\,(dI/dt) \). So the given emf and rate give the self-inductance \( L \). A coil carrying a steady current \( I \) stores the energy \( U=\frac{1}{2}LI^{2} \). This is the work done against the back emf while the current builds up.
\( \varepsilon=0.5\ \mathrm{V} \)
\( dI/dt=10\ \mathrm{A\,s^{-1}} \)
\( U=0.4\ \mathrm{J} \)
\( \begin{aligned} L &= \frac{\varepsilon}{dI/dt} \\ &= \frac{0.5}{10} \\ &= 0.05\ \mathrm{H} \end{aligned} \)
\( \begin{aligned} I^{2} &= \frac{2U}{L} \\ &= \frac{2\times0.4}{0.05} \\ &= 16\ \mathrm{A^{2}} \\ I &= 4\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “… When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. … the induced emf is given by”
\( \displaystyle \varepsilon=-\frac{d(N\Phi_B)}{dt} \)
\( \displaystyle \varepsilon=-L\frac{dI}{dt} \)\( \qquad (6.14) \)
Self-inductance – “Thus, the energy required to build up the current I is,”
\( \displaystyle W=\frac{1}{2}LI^{2} \)\( \qquad (6.17) \)