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Q. The energy stored in the magnetic field if current of $5\, A$ produces a magnetic flux of $2 \times 10^{-3} Wb$ through a coil of 500 turns is

Electromagnetic Induction

Solution:

$N \phi =L I$
$L=\frac{N \phi}{I} =\frac{500 \times 2 \times 10^{-3}}{5}=0.2\, H.$ Energy stored,
$U_{m}=\frac{1}{2} L I^{2}=\frac{1}{2} \times 0.2 \times(5)^{2}=2.5\, J$