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Starting from the expression for theenergy W= 1/2 LI^(2), stored in a solenoid of self-inductance L to build up the current I, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and lengthl of the solenoid having n number of turns per unit length. Hence show that the energy density is given by B^(2)//2mu_(0). |
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Answer» Solution :GIVEN, Energy `W=1/2 LI^(2)` A SOLENOID having magnetic field B. Area A & length l and having N numbersof turns PER unit length. Self inductance of the solenoid : `L= mu_(0)n^(2)lA` `B=mu_(0)n I` `:. W=1/2 mu_(0)n^(2)LAI^(2)` `:.B^(2)=mu_(0)^(2)n^(2)I^(2)` `V=AL` (volume) `rArr W=(1)/(2mu_(0)) mu_(0)n^(2)lAI^(2)` `rArr W=(1)/(2 mu_(0))B^(2)V` Energy density `=W/V = (B^(2))/(2mu_(0))` |
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