A uniform straight horizontal conductor of mass `m` and length `l` is pivoted at one end on a smooth horizontal surface. A vertical magnetic field `B` is present. A charge `Q` is passed through conductor at once. Find the angular velocity gained by the rod.

Options

A. `(QB)/(2m)`
B. `(6QB)/(m)`
C. `(3QB)/(m)`
D. `(3)/(2)(QB)/(m)`

Answer

D

Explanation

Let at any time current is `I` , then `F=IlB`
`tau=F(l)/(2)=(Il^(2)B)/(2)`

Angular impulse= change in angular momentum
`implies inttaudt=I_(A)omega`
`(Bt^(2))/(2)intIdt=(ml^(2))/(3)omega`
`(B^(2)l)/(2)Q=(ml^(2))/(3)omegaimpliesomega(3)/(2)(BQ)/(m)`