A circular tube of small thickness having outer radius r = R/2 is fixed in gravity free space. Magnetic field of strength B is present perpendicular to plane of circular tube as shown in figure. A cha

Question

A circular tube of small thickness having outer radius r = R/2 is fixed in gravity free space. Magnetic field of strength B is present perpendicular to plane of circular tube as shown in figure. A charged particle (q, m) inside the tube located at r = R/2 is given tangential velocity v 0 ​ = m qBR ​ . The work done by friction long after motion is − 8 m K q 2 B 2 R 2 ​ . Find the value of K. [Inner surface of tube is rough].

TestHub · Accepted Answer

Friction stops reducing the speed when the contact force becomes zero.

$\frac{m v _{f}^{2}}{R /2} = q v_{f} B \Rightarrow v_{f} = \frac{qBR}{2 m}$

$W_{f} = \frac{1}{2} m (v_{f}^{2} - v_{0}^{2})$

$W_{f} = - \frac{3 q ^{2} B ^{2} R ^{2}}{8 m} \Rightarrow K = 3$

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