InterviewSolution
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InterviewSolution
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Obtain an expression for magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solonoid. (b) How does this magnetic energy compare with the elecrostatic energy strored in a capacitor ? |
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Answer» SOLUTION :(a) The magnetic energy is `U_(B) = (1)/(2) LI^(2)` As, `B = (mu_(0) NI)/(l)`, THEREFORE, `I = (B l)/(mu_(0) N) :. U_(B) = (1)/(2) L ((B l)/(mu_(0) N))^(2)` Using `L = (mu_(0) N^(2) A)/(l)` , we get, `U_(B) = (1)/(2) ((mu_(0) N^(2) A)/(l)) ((B l)/(mu_(0) N))^(2)` `U_(B) = (B^(2))/(2 mu_(0)) A l` (b) The volume that contains flux, `V = A xx l` `:.` Magnetic energy per unit volume, `u_(B) = (U_(B))/(V) = (B^(2))/(2 mu_(0))` It is known that electrostatic energy stored per unit volume in a parallel plate capacitor is `u_(E) = (1)/(2) in_(0) E^(2)` In both the cases. energy is DIRECTLY proportional to the square of the field strenth. Note that Eqna. (II) and (iii) have been derived for special cases -- a solenoid and a parallelplate capacitor respectively. But they are valid for any region of space in which a magnetic field `or//and` an electric field exists. |
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