A copper rod of mass m = 1.00 kg rests on two horizontal rails a distance L = 1.

Question

A copper rod of mass m = 1.00 kg rests on two horizontal rails a distance L = 1.02 m apart and carries a current of i = 52.0 A from one                    rail to the other. A top view and a side view are shown in the figure. The coefficient of static friction between rod and rails is ? = 0.690. What are                    the (a) magnitude and (b) angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?

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Explanation / Answer

f = IBxl

= IBlsin(theta)

so theta = 90 ( to minimise B)

a) f=IBl=0.69*9.8*1 = 6.762

B=6.762/(52*1.02) =0.127

B) angle = 0 with vertical

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