An inductor coil of self-inductance $10 ext{ H}$ carries a current of $1 ext{ A}$. The magnetic field ener

Question

An inductor coil of self-inductance $10 	ext{ H}$ carries a current of $1 	ext{ A}$. The magnetic field energy stored in the coil is:

Accepted Answer

The magnetic potential energy ($U$) stored in an inductor is given by the formula:
$$U = \frac{1}{2} L I^2$$

**Given:**
*   Self-inductance ($L$) = $10 	ext{ H}$
*   Current ($I$) = $1 	ext{ A}$

**Calculation:**
$$U = \frac{1}{2} 	imes 10 	imes (1)^2$$
$$U = 5 	imes 1$$
$$U = 5 	ext{ J}$$