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A thin non-conducting ring of mass m car...

A thin non-conducting ring of mass `m` carrying a charge `q` can freely rotate about its axis. At `t = 0`, the ring was at rest and no magnetic field was present. Then suddenly a magnetic field `B` was set perpendicular to the plane. Find the angular velocity acquired by the ring.

Text Solution

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Due to the sudden change of flux, an electric field is set up and the ring experiences an impulsive torque and suddenly acquires an angular velocity.
`epsi` (induced emf) `=-(dphi)/(dt)=-(d)/(dt)intvecB.dA`
Also `epsi=ointvecE.dvecl` where E is the induced electric field.
`ointvecE.dvecl=-(d)/(dt)intvecB.dvecAimpliesE.2pi=-(d)/(dt)(Bpir^(2))`
`impliesE=-(r)/(2)(dB)/(dt)`
Force experienced by the ring `=q|vecE|`
Torque experienced by thering
`tau=(qE)r=(qr^(2))/(2)(dB)/(dt)`
`:.` Angular impulse experienced by the ring
`=inttaudt=(qr^(2))/(2)int(dB)/(dt)dt=qr^(2)(B)/(2)`
Also angular impulse acquired `=lomega` where l is moment of inertia of the ring about its axis `=mr^(2):.mr^(2)omega=qr^(2)B//2`
`implies` Angular velocity acquired by the ring `omega=qB//2m`
Note Now The student can now attempt Section - A from exercise.
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