Question
Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity $\omega.$ Find the force.

Answer


$\text{e}=\text{Bvl}=\frac{\text{B}\times\text{a}\times\omega\times\text{a}}{2}$
$\text{i}=\frac{\text{Ba}^2\omega}{2\text{R}}$
$\text{F}=\text{ilB}=\frac{\text{B}\text{a}^2\omega}{2\text{R}}\times\text{a}\times\text{B}=\frac{\text{B}^2\text{a}^2\omega}{2\text{R}}$ towards right of OA.

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