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Alternating Current question

2023 · Q151
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Alternating Current question

2023 · Q151

NEETPhysicsAlternating CurrentMCQ+4 / −1

The magnetic energy stored in an inductor of inductance 4 μH4 ~\mu \mathrm{H}4 μH carrying a current of 2 A2 \mathrm{~A}2 A is :

  1. A
    4 mJ4 \mathrm{~mJ}4 mJ
  2. B
    8 mJ8 \mathrm{~mJ}8 mJ
  3. C
    8 μJ8 ~\mu \mathrm{J}8 μJ
  4. D
    4 μJ4 ~\mu \mathrm{J}4 μJ
View written solutionFree

Correct answer: C

The formula for calculating the magnetic energy stored in an inductor is given by:

$$E = \frac{1}{2} L I^2$$

where:

  • $E$ is the energy stored (in joules, J),
  • $L$ is the inductance of the inductor (in henrys, H),
  • $I$ is the current flowing through the inductor (in amperes, A).

Plugging the given values into the formula:

$$E = \frac{1}{2} \times 4 \mu H \times (2 A)^2$$

$$E =\frac{1}{2} \times 4 \times 10^{-6} H \times 4 A^2$$

$$E = 2 \times 10^{-6} H \cdot 4$$

$$E = 8 \times 10^{-6} J$$

Hence, the energy stored in the inductor is $8 \mu J$.

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