(B) The self-inductance $L$ of the coil is given by the formula $L = \frac{N \phi}{I}$.Given: $N = 400$, $\phi = 3 \times 10^{-3}$ $Wb$, and $I = 4$ $

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(B) The self-inductance $L$ of the coil is given by the formula $L = \frac{N \phi}{I}$.Given: $N = 400$, $\phi = 3 \times 10^{-3}$ $Wb$, and $I = 4$ $A$.Calculating $L$: $L = \frac{400 \times 3 \times 10^{-3}}{4} = 100 \times 3 \times 10^{-3} = 0.3$ $H$.The energy $U$ stored in the coil is given by $U = \frac{1}{2} L I^2$.Substituting the values: $U = \frac{1}{2} \times 0.3 \times (4)^2 = 0.15 \times 16 = 2.4$ $J$.

Class 12 Physics Electromagnetic Induction Energy Stored in Inductor

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If a current of $4$ $A$ produces a magnetic flux of $3 \times 10^{-3}$ $Wb$ through a coil of $400$ turns, the energy stored in the coil will be: (in $J$)

MHT CET 2026 All MHT CET

A
$1.2$
B
$2.4$
C
$24$
D
$240$

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Electromagnetic Induction

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Energy Stored in Inductor

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If a current of $4$ $A$ produces a magnetic flux of $3 \times 10^{-3}$ $Wb$ through a coil of $400$ turns, the energy stored in the coil will be: (in $J$)

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