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Question

Figure shows a rectangular loop conducting $$PQRS$$ in which the arm $$PQ$$ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm $$PQ$$ is moved with the velocity $$v$$ towards the arm $$RS$$. Assuming that the arms $$QR, RS$$ and $$SP$$ have negligible resistance and the moving arm $$PQ$$ has the resistance $$r$$, obtain the expression for
The force 


Solution

The force required to keep the arm $$PQ$$ in 
Constant motion
$$F=BI\ l=B\left(\dfrac{Blv}{r}\right)l=\dfrac{B^{2}l^{2}v}{r}$$

Physics

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