1.

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. `3.125xx10^(-3)J`B. `6.25xx10^(-4)J`C. `1.25xx10^(-2)J`D. `5.0x10^(-4)J`

Answer» Correct Answer - A
Speed of the loop should be
`v=(l)/(t)=(0.5)/(2)=0.25m//s`
Induced emf, `eBvl=(1.0)(1.0)(0.25)(0.5)`
`=0.125V`
`:.` Current in the loop `i=(e)/(R )=(0.125)/(10)`
`=1.25xx10^(-2)A` The magnetic force on the left arm due to the magnetic field is
`F_(m)=ilB=(1.25xx10^(-2)(0.5)(1.0)`
`=6.25xx10^(-3)N`
to pull the loop uniform an external force of `6.25xx10^(-3)N` towards right must be applied.
`:. W=(6.25xx10^(-3N)(0.5m)=3.125xx10^(-3)J`


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