1.

A rod OA of length l is rotating (about end O) over a conducting ring in crossed magnetic field B with constant angular velocity omega as shown in figure.

Answer»

Current flowing through the rod is `(3B omega l^(2))/(4R)`
Magnetic FORCE acting on the rod is `(3B^(2)omega l^(3))/(4R)`
Torque due to magnetic force acting on the rod is `(3B^(2)omega l^(4))/(8R)`
Magnitude of external force that acts perpendicularly at the end of the rod of maintain the constant angular speed is `(3B^(2)omega l^(3))/(8R)`

Solution :`I=(epsilon)/((2R)/(3))=(3epsilon)/(2R)=(3)/(2R)XX(1)/(2)B omega l^(2)=(3B omega l^(2))/(4R)`
Magnetic force
`F=(3B omega l^(2))/(4R)xx l xx B=(3B^(2)omega l^(3))/(4R)`
Torque, `tau =(3B^(2)omega l^(3))/(4R)xx(l)/(2)=(3B^(2)omega l^(4))/(8R)`
`therefore` Force to be applied at the end `= (3B^(2)omega l^(3))/(8R)`


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