Question

A wire of length `1 m` is moving at a speed of `2 ms^(-1)` perpendicular to its length and a homogeneous magnetic field of `0.5T`. The ends of the wire are joined to a circuit of resistance `6 Omega`. The rate at which work is being done to keep the wire moving at constant speed is
A. 1W
B. `(1)/(3)W`
C. `(1)/(6)W`
D. `(1)/(12)W`

Answer 1

Correct Answer - C
Rate of doing work `=(W)/(t)=P`
But, P = Fv
Also, `F = Bil=B((Bvl)/(R))l`
`[because"induced current, I"=(Bvl)/(R)]`
`therefore "Power, P= B"((Bvl)/(R))lv=(B^(2)v^(2)l^(2))/(R)`
`=((0.5)^(2)xx(2)^(2)xx(1)^(2))/(6)=(1)/(6)W`