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A conducting rod of length 2l is rotating with constant angular speed omega about is perpendicular bisector. A uniform magnetic field vec(B) exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is |
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Answer» `(1)/(2)B omega l^(2)` `e_(1)=(1)/(2)Bl^(2)omega` and between centre and other end `e_(2)=-(1)/(2)Bl^(2)omega`. Therefore between two ENDS `e=e_(1)+e_(2)=0` `therefore` Between two ends of the rod, e.m.f. induced = 0. |
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