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A rectangular loop with a sliding connector of length `10 cm` is situated in uniform magnetic field perpendicular to plane of loop. The magnetic induction is `0.1` tesla and resistance of connector `(R )` is `1 ohm`. The sides `AB` and `CD` have resistances `2 ohm` and `3 ohm` respectively. Find the current in the connector during its motion with constant velocity one A. `(1)/(242)A`B. `(1)/(220)A`C. `(1)/(55)A`D. `(1)/(440)A` |
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Answer» Correct Answer - B `e=Bvl=(0.1)(1)(0.1)=(1)/(100)V` Net resistance `=1+(2xx3)/(2+3)=(11)/(5)Omega` `therefore` Current, `i=((1//100)/(11//5))=(1)/(220)A` |
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