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Charge 'q' is uniformly distributed alon...


Charge 'q' is uniformly distributed along the length of a non-coducting circular ring of mass m and radius 'a'. The ring is placed concentrically magnetic field B=kt (where k is constant) is existing perpendicular to the plane of circular region of radius 'R' as shown. The minimum coefficient of friction between the ring and the surface required to keep the ring stationary is

A

`(2kqa)/(mg)`

B

`(kqa)/(mg)`

C

`(kqa)/(2mg)`

D

`(kqa)/(4mg)`

Text Solution

Verified by Experts

The correct Answer is:
C

`E2pir=(d)/(dt)(ktpir^(2))`
`E=(kr)/(2),rleR`
at r=a
`E=(ka)/(2)`
To keep the ring stationary qEa`=mumga`
`(qka)/(2)=mumg`
`thereforemu=(kqa)/(2mg)`
`mu_(min)=(kqa)/(2mg)`
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