A steady current of 4 A flows through a coil. For this current, the product \( N\Phi_B \) for the coil is 0.06 Wb, where N is the number of turns and \( \Phi_B \) the magnetic flux through each turn. The magnetic energy stored in the coil is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.12 J
The self-inductance of a coil is its flux linkage per unit current, \( L=N\Phi_B/I \). Work has to be done against the back emf to build up a current \( I \) in the coil. This work is stored as magnetic potential energy \( W=\frac{1}{2}LI^{2} \).
The self-inductance of the coil is:
\( \begin{aligned} L &= \frac{N\Phi_B}{I} \\ &= \frac{0.06\,\mathrm{Wb}}{4\,\mathrm{A}} \\ &= 0.015\,\mathrm{H} \end{aligned} \)
The energy stored for a current of 4 A is:
\( \begin{aligned} W &= \frac{1}{2}LI^{2} \\ &= \frac{1}{2}\times0.015\,\mathrm{H}\times(4\,\mathrm{A})^{2} \\ &= 0.12\,\mathrm{J} \end{aligned} \)