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

Options

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

Answer

A

Explanation

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`