InterviewSolution
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InterviewSolution
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Figure show a square loop of side `0.5 m` and resistance `10Omega`. The magnetic field has a magnitude `B=1.0T`. The work done in pulling the loop out of the field slowly and uniformly in `2.0s` is A. (a) `3.125 xx 10^(-3)J`B. (b) `6.25 xx 10^(-4)J`C. ( c) `1.25 xx 10^(-2)J`D. (d) `5.0 xx 10^(-4)J` |
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Answer» Correct Answer - A (a) Speed of the loop should be `v = (l)/(t) = (0.5)/(2) = 0.25 m s^(-1)` Induced emf, `e = Bvl = (1.0)(0.25)(0.5) = 0.125 V` `:.` Current in the loop, `I = (e)/(R ) = (0.125)/(10)` `=1.25 xx 10^(-2)A` The magnetic force on the left arm due to the magnetic field is `F_(m) = ilB = (1.25 xx 10^(-2))(0.5)(1.0)` `=6.25 xx 10^(-3) N` To pull the loop uniformly an external force of `6.25 xx 10^(-3)` `N` towards right must be applied. `:.` `W = (6.25 xx 10^(-3) N)(0.5m) = 3.125 xx 10^(-3)J` |
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