question_answer

A rod OA of length \[l\] is rotating (about end 0)  over a conducting ring in crossed magnetic field B with constant angular velocity co as shown in  figure

A) Current flowing through the rod is\[\frac{B\omega \,{{l}^{2}}}{R}\]

B) Magnetic force acting on the rod is \[\frac{3{{B}^{2}}\omega \,{{l}^{3}}}{4R}\]

C) Torque due to magnetic force acting on the rod is\[\frac{3{{B}^{2}}\omega \,{{l}^{4}}}{8R}\]

D) Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is \[\frac{3{{B}^{2}}\omega \,{{l}^{3}}}{5R}\].            

Correct Answer: C

Solution :

[c] \[I=\frac{\varepsilon }{\frac{2R}{3}}=\frac{3\varepsilon }{2R}\]             \[=\frac{3}{2R}\times \frac{1}{2}B\omega {{l}^{2}}\] \[=\frac{3B\omega {{l}^{2}}}{4R}\] Magnetic force \[F=\frac{3B\omega {{l}^{2}}}{4R}\times l\times B=\frac{3{{B}^{2}}\omega {{l}^{3}}}{4R}\] \[\tau =\frac{3{{B}^{2}}\omega {{l}^{3}}}{4R}\times \frac{l}{2}=\frac{3{{B}^{2}}\omega {{l}^{4}}}{8R}\] \[\therefore \]    Force to be applied at the end\[=\frac{3{{B}^{2}}\omega {{l}^{3}}}{8R}\].