A copper rod of mass 0.6 kg rests on two rails 1.3 m apart and carries a current of 31 A from one rail to the other. The coefficient of static friction is 0.3. What is the magnitude of the least magnetic field that would cause the bar to slide?
f=(Fn)(u) where f is the frictional force, Fn is the normal force, and u is mu.
The Attempt at a Solution
I set up a free body diagram and found the equation: F-(Fn)(u)=ma. In the y direction Fn=mg=(0.6)(9.8)= 5.88N. Then I plugged that into my first equation for the x direction and got F= 0.6a + 5.88. I know that F=ILB, so I plugged that in and solved for B: B= (.6a + 1.764)/40.3. Am I on the right track? If I am then where do I get the acceleration of the copper rod from? Thanks