Electromagnetic Induction

JEE Main level, numerical answer, medium

105 of 153

Question 105 of 153

A coil has an inductance of 2H2H and resistance of 4Ω4\Omega. A 10V10V is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be _____ ×10−2J\times {10}^{-2}J

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Answer: 625

The data given is
L=2HR=4ΩV=10VL=2H R=4\Omega V=10V
Using Ohm's law, V=IRV=IR,
I=10V4Ω=52AI=\dfrac{10V}{4\Omega }=\dfrac{5}{2}A
The energy of an inductor is given by
E=12LI2=12×2×(254)=625×10−2JE=\dfrac{1}{2}L{I}^{2}=\dfrac{1}{2}\times 2\times \left(\dfrac{25}{4}\right)=625\times {10}^{-2}J

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