Question 269: Electromagnetic Induction

Question

The magnetic energy stored in an inductor of inductance $5 \mu \mathrm{H}$ carrying a current of 2 A is

Accepted Answer

The magnetic energy stored in an inductor can be calculated using the formula:
$$ E = \frac{1}{2} L I^2 $$
where:
$ E $ is the energy stored in joules,
$ L $ is the inductance in henries,
$ I $ is the current in amperes.
Given:
$ L = 5 \, \mu\mathrm{H} = 5 	imes 10^{-6} \, \mathrm{H} $,
$ I = 2 \, \mathrm{A} $.
Substitute these values into the formula:
$$ E = \frac{1}{2} 	imes 5 	imes 10^{-6} 	imes (2)^2 $$
Calculate the energy:
$$ E = \frac{1}{2} 	imes 5 	imes 10^{-6} 	imes 4 $$
$$ E = 10 	imes 10^{-6} \, \mathrm{J} = 10 \, \mu \mathrm{J} $$
Thus, the magnetic energy stored in the inductor is $ 10 \, \mu \mathrm{J} $.
The correct option is
Option C
$ 10 \, \mu \mathrm{J} $.

Answer

10 mJ

Answer

5 mJ

Answer

$10 \mu \mathrm{~J}$

Answer

$5 \mu \mathrm{~J}$

Electromagnetic Induction

The magnetic energy stored in an inductor of inductance 5 \mu \mathrm{H} carrying a current of 2 A is

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