A nonconducting ring of uniform mass m, radius b and uniform linear charge density `lambda` is suspended as shown in figure in a gravity free space. There is uniform coaxial magnetic field `B_0`, pointing up in a circular region of radius 'a' `(lt b)`. Now if this field is switched off, then:-

Options

A. a. There will be induced electric field on periphery of ring, in anticlockwise sense when seen from above
B. b. Induced electric field imparts angular momentum of magnitude `lambda pia^(2)b B_(o)`
C. c. Final angular velocity of ring will be more if time taken to switch of the field `B_(o)` is small
D. d. Final angular velocity will always be independent of time taken to switch off the field `B_(o)`

Explanation

Changing magnetic field (at switching off `B_(0)` to zero) induced electric field in such a way to restore the upwards flux, hence anticlockwise (E) as seen from above
` intvec(E).vec(dl)=-(dphi)/(dt)=-pia^(2)(dB)/(dt)=intEdl`
There is force on small element dQ of ring, tangentially Now this force produces torque about axis of ring to rotate in anticlockwise sense, so
`tau=int dQExxb=intlambdadlEb=lambdabintEdl=lambdabpia^(2)(dB)/(dt)`
So impulse or torque
`int taudt=lambdab pia^(2)int_(B_(0))^(2)dB=int taudt=lambdabpi a^(2)B_(0)`
`L_(f)-L_(i)=DeltaL=inttau dt =lambda b pia^(2)B_(0) =Iomega` (in magnitude)
It is indpendent of time taken `I omega_(f)-Iomega_(i)=lambdab pia^(2)B_(0)`
where I is moment of inertia
So, `omega_(f)=(lambdab pi a^(2)B_(0))/(mR^(2))`