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A horizontal tube of small thickness having inner radius r=R//2 is placed in gravity free space. Magnet field of strength B is present perpendicular to plane of circular tube. A charged particle (q.m) inside, the tube is given tangential velocity v_(@)=(qBR)/m. The work done by friction long after motion is -(Kq^(2)B^(2)R^(2))/(8m). Find the value of K. [inner surface of tube is rough] |
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Answer» `W_("friction")=0-1/2m[(qBR)/m]^(2)` `W_("friction")=-1/2(Q^(2)B^(2)R^(2))/m=(-Kq^(2)B^(2)R^(2))/(8M)` `K=4` |
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