Saved Bookmarks
| 1. |
A non-conducting ring of radius r has charge per unit length `lambda`. A magnetic field perpendicular to plane of the ring changes at rate Db/dt. Torque experienced by the ring is A. `lambdapir^(3)(dB)/(dt)`B. `lambda2pir^(3)(dB)/(dt)`C. `lambda^(2)(2pir)^(2)(dB)/(dt)`D. zero |
|
Answer» Correct Answer - A `e= (dphi)/(dt)= pir^(2)(dB)/(dt)rArr E(2 pi r) = pir^(2)(dB)/(dt)` `E= (1)/(2) r (dB)/(dt)rArr dF = lambdadSE` `rArr " "d tau= rlambdaDSE rArr tau = lambdarEintdS = lambdarE2 pi r` |
|