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(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? |
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Answer» SOLUTION :When a current is established in a solenoid (coil), work has to bedone against the BACK emf. This work done is stored in the form of magnetic energy in the coil. For a current I in the coil, the rate of work done (power) is : `P = (DW)/(dt) = |EPSILON|` I (since power `P=VI` ) But we know that , `epsilon = L (dI)/(dt)` Therefore, `(dW)/(dt) = L I (dI)/(dt) rArr dW = LI dI` Therefore, the total work done in establishing a current I is given by `W = int d W = int_(0)^(1) L I d I = ""_(1//2)LI^(2)` This work is stored in the coil in the form of magnetic potential energy, `U= 1/2 L I^(2)` |
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