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The current in the inner coil is I = 2t^(2). Find the heat developed in the outer coil between t =0 and t seconds. The resistance of the outer coil is R and take b gt gt a. |
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Answer» Solution :Let the current I be in the OUTER coil. The field at CENTRE `B = (mu_(0)^(I))` The flux through the inner coil = `(mu_(0)Ipia^(2))/(2b)` The induced emf produced in the outer coil `epsilon = -(dphi)/(dt)` `(mu_(0)OUA^(2))/(2b)(d)/(dt)(2t^(2)) = (2mu_(0)PIA^(2)t)/(b)` Current induced in the outer coil = `(epsilon)/(R ) = (2mu_(0)pia^(2)t)/(bR)` Heat developed in the outer coil = `int_(0)^(t)l^(2)"Rdt"= int_(0)^(t)(4mu_(0)^(2)PI^(2)a^(4)t^(2)Rdt)/(b^(2)R^(2))=(4mu_(0)^(2)pi^(2)a^(4))/(b^(2)R) (t^(3))/(3)` |
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