InterviewSolution
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Derive an expression for the energy stored in the magnetic field of an inductor. OR Derive an expression for the electrical work done in establishing a steady current in a coil of self inductance L. |
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Answer» Consider an inductor of sell inductance L connected in a circuit When the circuit is dosed, the current in the circuit increases and so does the magnetic flux linked with the coiL At any instant the magnitude of the induced emf is e = L \(\cfrac{di}{dt}\) The power consumed in the inductor is P = ei = L \(\cfrac{di}{dt}\) . i [Alternatively, the work done in moving a charge dq against this emf e is dw = edq = L \(\cfrac{di}{dt}\). dq = Li ∙ di (∵ \(\cfrac{dq}{dt}\) = i) This work done is stored in the magnetic field of the inductor. dw = du.] The total energy stored In the magnetic field when the current increases from 0 to I In a time interval from 0 to t can be determined by integrating this expression : Um = ∫0t Pdt = ∫0I Lidi = L ∫0I idi = \(\cfrac12\) LI2 which is the required expression for the stored magnetic energy. [Note: Compare this with the electric energy stored in a capacitor, Ue = \(\cfrac12\) CV2] |
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