A 2.4 kg copper rod rests on two horizontal rails (see figure below) 2.4 m apart

Question

A 2.4 kg copper rod rests on two horizontal rails (see figure below) 2.4 m apart and carries a current of 40 A from one rail to the other. The coefficient of static friction between rod and rails is 0.42. What is the smallest magnetic field (not necessarily vertical) that would cause the rod to slide? (Based on the bottom picture, define to the right as the +x-direction and up as the + indirection. Assume the current in the bottom picture is into the page and the magnetic force will cause the rod will slide to the right.)

Explanation / Answer

Here ,

mass of rod , m = 2.4 Kg

coefficient of static friction , u = 0.42

current , I = 40 A

length , L = 2.4 m

for moving the rod

frictional force = magnetic force

u * mg = B*I*L

0.42 * 2.4 * 9.8 = B * 40 * 2.4

B = 0.103 T

the magnetic field is 0.103 T

as B = iL X B

the direction of field must be out of the page

theta = 90 degree

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