A circular coil (500 turns, radius = 0.06 m) is rotating in a uniform magnetic f

Question

A circular coil (500 turns, radius = 0.06 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.010 s, the normal makes an angle of phi = 45° with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.100 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

Explanation / Answer

E = - N (dF/dt) F = BAcos(wt) take derivative of F with respect to t you get dF = - BAwsin(wt) E = + NBAwsin(wt) given : E= 0.100 V N=500 Area=A= p r ^2 B = ? w = convert-->(1/8)rev/.010sec-->12.5rev/sec… = 2p-->12.5*2p-->25prad/sec t= .010sec plug everything in and solve for B: E = NBAwsin(wt) B= E/(NAwsin(wt) B=0.100 V / [(500)(p (0.06 m) ^2)( 25p rad/sec)sin((25p rad/sec)*.010sec)]

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