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A wire frame in the form of a part of circle ( sector ) of radius 1 and resistance R is free to rotate about on axis passing through O and perpendicular to plane of paper as shown in the figure. The angle of the sector is (pi)/(4) and it is rotating with constant angular velocity omega as shown. Above line PQ uniform magnetic field of magnitude B exists in the direction perpendicular to plane of paper. In region I field is outward while in region II, field is inward. Based on above information, answer the following questions : Total thermal energy dissipated in one cycle is |
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Answer» `(B^(2)omegapl^(4))/(16R)` `e_(1)=` induced `emf=(Bomegal^(2))/(2)` Thermal energy DISSIPATED in time `t_(1)` `H_(1)=(e_(1)^(2))/(R)t_(1)=(B^(2)omegapil^(4))/(16R)` SIMILARLY , thermal energy developed when the frame moves fromregion I to region II is , `H_(2)=(B^(2)omegapil^(4))/(4R)` Thermal energy will also develop when the loop comes out from magnetic field and is given by, `H_(3)=(B^(2)omegapil^(4))/(16R)` So, total thermal energy developed in one revolution is, `H=H_(1)+H_(2)+H_(3)=(3B^(2)omegapil^(4))/(8R)` Average power produced, `P_(av)=(H)/(T)=(3B^(2)omega^(2)l^(4))/(16R)` |
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