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A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is `15ms^(-1)` and is directed perpendicular to the field. The emf induced between the two ends of the conductor isA. 0.10 VB. 0.15 VC. 1.50 VD. 15 V |
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Answer» Correct Answer - B Given, length of conductor, l = 0.1m Magnetic field, B = 0.1 T Veocity of conductor, `v=15ms^(-1)` The angle between v and B is `90^(@)`. When v and B are mutually perpendicular, then emf (induced) is given by `epsilon=vBl=15xx0.1xx0.1=(15)/(100)=0.15V` |
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