Question

The magnetic energy stored in an inductor of inductance 14 µH carrying a current of 2 A is (in µJ)

Answer 1

28 µJ

Answer 2

The magnetic energy ($U$) stored in an inductor is given by the formula:

$U = \frac{1}{2}LI^2$

where $L$ is the inductance and $I$ is the current flowing through the inductor.

Given values: Inductance, $L = 14 \, \mu H = 14 \times 10^{-6} \, H$ Current, $I = 2 \, A$

Substitute the values into the formula: $U = \frac{1}{2} \times (14 \times 10^{-6} \, H) \times (2 \, A)^2$ $U = \frac{1}{2} \times 14 \times 10^{-6} \times 4$ $U = 7 \times 10^{-6} \times 4$ $U = 28 \times 10^{-6} \, J$

Since $1 \, \mu J = 10^{-6} \, J$, the energy in microjoules is: $U = 28 \, \mu J$