Practice portal › Electromagnetic Induction › Self-Inductance
Asked in GUJCET 2007 · Energy stored in an inductor
Given: L = 50 mH = 50×10⁻³ = 0.05 H, I = 2 A.
Idea: the work done in building up the current against the back emf is stored in the magnetic field of the coil: U = 1/2LI².
U = 1/2×0.05×(2)² = 1/2×0.05×4.
U = 0.1 J.
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