Question: 1 The magnetic energy stored in an inductor of inductance 4 \mu \mathrm{H} carrying a current of 2 A is :
(1) 8 mJ
(2) 8 \mu \mathrm{~J}
(3) 4 \mu \mathrm{~J}
(4) 4 mJ
Answer: Option (2)
Explanation:
The magnetic energy stored in an inductor is given by the formula U=\frac{1}{2} L I^{2}.
Here, the inductance is L=4 \mu \mathrm{H}=4 \times 10^{-6} \mathrm{H}
and the current is I=2 \mathrm{A}.
Substituting the values, the energy stored is
U=\frac{1}{2} \times 4 \times 10^{-6} \times (2)^{2}.
On simplifying, we get U=\frac{1}{2} \times 4 \times 10^{-6} \times 4.
This gives U=8 \times 10^{-6} \mathrm{J}.
Therefore, the magnetic energy stored in the inductor is 8 \mu \mathrm{J}.