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A current flowing in the winding of a long straight solenoid is increased at a sufficiently slow rate. Demonstrate that the rate at which the enrgy of the magnetic field in the solenoid increases is equal to the flux of the Poynting vector across the lateral surface of the solenoid. |
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Answer» Solution :Within the solenoid `B = mu_(0)n I` and the rate of CHANGE of magnetic energy `=W_(m) = (d)/(dt) ((1)/(2)mu_(0)n^(2)I^(2)piR^(2)l) = mu_(0)n^(2)piR^(2)l I dotI` where `R =` radius of CROSS section of the solenoid `l =` length Also `H = B//mu_(0) = nI` along the axis within teh solenoid. By Faraday's law, the induced electirc field id `E_(theta) 2pir = pir^(2)dotB = pir^(2)mu_(0)n dotI` or `E_(theta) = (1)/(2)mu_(0) n dotI r` so at the edeg `E_(theta)(R) = (1)/(2)mu_(0)n dotIR` (circuital) Then `S_(r ) = E_(theta)H_(z)` (radially inward) and `dotW_(m) = (1)/(2)mu_(0)n^(2)I dotIR XX 2piRl = mu_(0)n^(2)piR^(2) lII` as before. |
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