InterviewSolution
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InterviewSolution
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(a) Obtain the expression for the magnetic energy stored in a solenoid, due to the current I flowing in it, in terms of magnetic field B, area of cross-section A and length l of the solenoid. (b) How is this magnetic energy per unit volume compared with the electrostatic energy per unit volume stored in a parallel plate capacitor ? |
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Answer» Solution :(a) We know that a solenoid having self-inductance L carrying current I has magnetic energy stored in it whose value is given by `U_(B) = 1/2 Ll^(2)` As magnetic field developed INSIDE the solenoid is `B = mu_(0)n^(2)I` hence `I=B/(mu_(0)n)`. Moreover self-indutance of solenoid coil `L = mu_(0)n^(2)Al`. Substituting these values, we have `U_(B) = 1/2 (mu_(0)n^(2)Al)(B/(mu_(0) n))^(2) = 1/(2mu_(0)).AlB^(2)` (b) As V = A.l is the volume of solenoid coil, hence magnetic energy stored per unit volume in a current carrying solenoid is given by `u_(B) = (U_(B))/V(1/(2mu_(0))AlB^(2))/(Al) = B^(2)/(2mu_(0))` The result compares with the ELECTROSTATIC energy per unit volume stored in a PARALLEL plate capacitor `u_(B)=1/2=in_(0)E^(2)` in the sense that in both cases energy per unit volume is proportional to square of electric/magnetic field. |
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