A magnetic field B is confined to a region `r le` a and points out of the paper (the z-axis), r = 0 being the center of the circular region. A charged ring (charge = Q) of radius b and mass m lie in the x-y plane with its center at origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time `Delta t`. Find the angular velocity `omega` of the ring after the field vanishes.

Options

A. `(qBa^(2))/(2mb)`
B. `(qBa)/(2mb^(2))`
C. `(2b^(2))/(qBa^(2))`
D. `(qb^(2))/(2Ba^(2))`

Answer

D

Explanation

Let E is the electric field generated around the charged ring of radius of b, then,
`epsi=(dphi)/(dt)`
`ointvec(E)*dvec(l)=(Bpia^(2))/(Deltat)`
`or Eb=(Ba^(2))/(2(Deltat))" n "......(i)`
Torque acting on the ring
`tau=bxx"forces"=bqE`
`=(qBa^(2))/(2(Deltat)) " "` [ Using (i)]
If `DeltaL` is change in angular momentum of the chorged ring, then,
`tau=(DeltaL)/(Deltat)=(L_(2)-L_(1))/(Deltat)`
`:. L_(2)-L_(1)=tau(Deltat)`
`=(qBa^(2)Deltat)/(2Deltat)=(qBa^(2))/(2)`
As initial angular momentum, `L_(1)=0`
`:. L_(2)=(qBa^(2))/(2)=I omega=mb^(2)omega :. omega=(qBa^(2))/(2mb^(2))`