If current of 4 A produces magnetic flux of \(3 \times 10^{-3} \mathrm{~Wb}\) through a coil of 400 turns, the energy stored in the coil will be

\(\begin{array}{ll} & \text { Given } \mathrm{I}=4 \mathrm{~A}, \mathrm{~N}=400 \text { and } \\ & \phi=3 \times 10^{-3} \mathrm{~Wb} \\ & \mathrm{E}=\frac{\mathrm{LdI}}{\mathrm{dt}} \cdot \text { and } \mathrm{e}=\frac{\mathrm{Nd} \phi}{\mathrm{dt}} \\ \Rightarrow \quad & \frac{\mathrm{Nd} \phi}{\mathrm{dt}}=\frac{\mathrm{LdI}}{\mathrm{dt}} \\ \therefore \quad & \mathrm{N} \phi=\mathrm{LI} \\ & \mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{400 \times 3 \times 10^{-3}}{4}=0.3 \mathrm{H} \\ \therefore \quad & \text { Energy stored in the coil }=\frac{1}{2} \mathrm{LI}^2 \\ = & \frac{1}{2} \times 0.3 \times 16 \\ = & 2.4 \mathrm{~J}\end{array}\)