A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conduc

Question

A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod has a weight mg and it is falling with an instantaneous speed v. There is a constant magnetic field B which is perpendicular to the paper and directed into the paper. Find the direction of the induced current through the resistor It. A to B B to A 0 What is the magnitude of the induced current? If B = 3.6T, d = 7m, m = 4.7kg, R = 8.2 ohm, and g = 9.8 m/s2 . find the terminal velocity. (When the terminal velocity is readied, there is no net force on the rod, so the magnetic force is equal and opposite to the weight of the rod.) Answer in units of m/s.

Explanation / Answer

   a.   The direction of currentis from B to A so asit apposes the increase in magnetic flux.    b.   InducedEMF   E   =   B * v *d * sin                                  =   B* v *d            (as   =   900)       Inducedcurrent   I   =   E /R                                     =   B* v * d / R    c.   When terminal speed isreached,   mg   =   F       m *g   =   B * I * d * sin900       4.7 *9.8   =   B * (B * v * d / R) *d              46.06     =   3.62* v * 72 / 8.2       speed ofrod   v   =   46.06 *8.2 / 635.04                                  =   0.595   m/s       4.7 *9.8   =   B * (B * v * d / R) *d              46.06     =   3.62* v * 72 / 8.2       speed ofrod   v   =   46.06 *8.2 / 635.04                                  =   0.595   m/s

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