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

Options

A. `(1)/(12)W`
B. `(1)/(6)W`
C. `(1)/(3)W`
D. 1W

Answer

B

Explanation

Rate of work =
`(W)/(t)=P=Fv`, also `F=Bil=B((Bvl)/(R ))l`
`rArr P=(B^(2)v^(2)l^(2))/(R )=((0.5)^(2)xx(2)^(2)xx(1)^(2))/(6)=(1)/(6)W` [Here W = Watts]