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A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s_(1) in a uniform horizontal magnetic field of magnitude 3.0 xx 10^(2)T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 102Omega, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? |
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Answer» SOLUTION :Here RADIUS of coil R = 8.0 cm = 0.08 m, hence area of coil `A = piR^(2) =pi xx (0.08)^(-2) = 0.02 m^(2)`, number of turns in the coil N = 20, angular speed of rotation `omega = 50s^(-1)` and magnetic field `B = 3.0 xx 10^(2)T`. `therefere` Maximum induced emf in the coil `lvarepsilon_(MAX) = NBA omega` `= 20 xx 3 xx 10^(-2) xx 0.02 xx 50 = 0.6 V` Average induced emf `vareppsilon_(av)` = average value of `NBA omegasin omegat` for ONE cycle = `N B A omega` [average value of sin t for one cycle] =0. As resistance `R = 10Omega`, hence maximum induced current `I_(max) = (varepsilon_(max))/R = 0.6/10 A = 0.06A` `therefere` Average power loss due to Joule.s heating `P_(av) = (varepsilon_(max).I_(max))/2=(0.6 xx 0.06)/2 =0.018W. Source of power: The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) mustupply torque (and do work) to counter this torque in order to keep the coil rotating uniformly. Thus, the source of the power dissipated as heat in the coil is the external rotor. |
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