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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 3.0xx10^(-2)T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Omega, calculate teh maximum value of current in the coil Calculated the average power loss due to Joule heating. Where does this power come from? |
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Answer» Solution :For a coil ROTATING in a magnetic field, `varepsilon=-BANomegasinomegat` `varepsilon_(max)=-BANomega` (NUMERICALLY) Here `B=3xx10^(-2)T,A=pir^(2)=pixx(8XX10^(-2))^(2)` `N=20,omega=50" rad "s^(-1)` `varepsilon_(max)=20xx50xxpixx64xx10^(-4)xx3xx10^(-2)=0.603V` `varepsilon_("average")` over a cycle is zero `I_(max)=(varepsilon_(max))/(R)=(0.603)/(10)=0.0603A` `P_("average")=(1)/(2)varepsilon_(max)I_(max)=(1)/(2)xx0.603xx0.0603=0.018W` Source of power LOSS is the external rotor which provides the necessary torque. |
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