A small square coil of N turns has sides of length L and is mounted so that it c

Question

A small square coil of N turns has sides of length L and is mounted so that it can pivot freely about a horizontal axis PQ, parallel to one pair of sides of the coil, through its centre. The coil is situated between the poles of a magnet which produces a magnetic field of flux density B. The coil is mantained in a vertical plane by moving a rider of mass M along a horizontal beam attached to the coil. When a current I flows through the coil, equilbrium is restored by placing the rider at a distance x along the beam from the coil. Show that B is given by the expression B=Mgx/NIL2

Explanation / Answer

n turns, L=side square, surface area =L^2, volume =area*N summation of forces to do equilibrium: Mg=Felectric Mg= I * B* N* L^2 (since electrom motive force is I dot (L crossB) using right hand rule, and L^2 since two sides) PLZ RATE AT LEAST HELPFUL, thx

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