A 0.150-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.510 m apart. If the coefficient of kinetic friction between the rod and rails is 0.160, what vertical magnetic field is required to keep the rod moving at a constant speed?

B = 0.0307 T = 30.74 mT

Explanation:

Given

m = 0.150 kg

I = 15.0 A

d = 0.510 m

μk = 0.16

B = ?

Balancing the forces on the rod in the j direction

N - m*g = 0 ⇒ N = m*g

and in the i direction

I*d*B - μk*N = 0 ⇒ B = μk*N / (I*d)

⇒ B = μk*m*g / (I*d)

⇒ B = (0.16) * (0.150 kg) * (9.8 m/s²) / (15.0 A*0.510 m)

⇒ B = 0.0307 T = 30.74 mT