A 1.20 kg copper rod rests on two horizontal rails 0.78 m apart and carries a current of 45 A from one rail to the other. The coefficient of static friction between rod and rails is 0.61. What is the magnitude of the smallest magnetic field that puts the rod on the verge of sliding

B = 0.204T

Explanation:

To find the value of the magnetic force you use the following formula:

I: current of the copper rod = 45A

B: magnitude of the magnetic field

L: 0.78m

you assume that magnetic field B and current I are perpendicular between them.

The magnetic force must be, at least, equal to the friction force, that is:

M: mass of the rod = 1.20kg

μ: coefficient of static friction = 0.61

g: gravitational acceleration constant = 9.8m/s^2

By replacing the values of the parameters you obtain:

the answer is a. hope this .