InterviewSolution
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InterviewSolution
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A source of emf varepsilon is used to establish a current I through a coil of self-inductance L. Show that the work done by the source to build up the current I_(0) is 1/2 LI_(0)^(2). Or Obtain the expression for the magnetic energy stored in an ideal inductor of self-inductance L when a current I_(0) passes through it. |
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Answer» Solution :Consider a battery CONNECTED to an inductor L. As the battery cireuit is completed, current starts growing in inductor and consequently an induced EMF is set up in the reverse direction given by `|varepsilon| = L (dI)/(dt)` If the battery supplies a current I through the inductor for a small time dt, the elementary work done by emf source (battery) against the induced emf will be `dW = varepsilon dt = L (dI)/dt.I.dt = LIdI` Hence, the total amount of work done by emf source till the current increases from initial zero VALUE to its FINAL steady value `I_(0)`, will be `W = int_(0)^(I_(0)) LI dI = =[1/2LI_(0)^(2)]_(0)^(I_(0))=1/2L I_(0)^(2)` This work done is stored in the inductor in the form of its MAGNETIC energy. Thus, Energy stored in an inductor `U = W = 1/2LI_(0)^(2). |
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