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A current carrying ring, carrying a cons...

A current carrying ring, carrying a constant current `(2)/(pi) Amp`, radius 1m mass `(2)/(3)kg` having 10 windings is free to rotate about its tangential vertical axis. A uniform magnetic field of 1 tesla is applied perpendicular to its plane. How much minimum angular velocity (in rad/sec) should be given to the ring in the direction shown, so that it can rotate `270^(@)` in that direction. Write your answer in nearest single digit in rad/sec

Text Solution

Verified by Experts

The correct Answer is:
9

to reach `theta = 270^(@)`,It has to cross the potential energy barrier at `0 = 180^(2)` and to cross `theta = 180^(@)` angular velocity at `theta = 180^(@)` should be `0^(+)`
`k_(1) + U_(i) = K_(f) + U_(y)`
`(1)/(2)((3)/(2)MR^(2))omega^(2)+(-MiAB cos 0^(@))=0+(-NiAB cos 180^(@)) omega = sqrt(80) ~~ 9 rad//sec`.
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