# A copper rod weighing 0.14 kg, lying across two rails spaced 0.3 m from each other, carries a current of 50 A

**A copper rod weighing 0.14 kg, lying across two rails spaced 0.3 m from each other, carries a current of 50 A. The coefficient of sliding friction of the rod along the rails is 0.6. What is the minimum amount of magnetic flux density that can cause the rod to slide, and what is the direction of the current?**

Let’s write it down briefly:

copper rod.

m = 0.14 kg;

L = 0.3 meters;

I = 50 A;

k = 0.6.

Find:

B -?

Let’s start by determining what kind of force acts on the rod and how to find this force:

1. Friction force as k * m * g;

2. Ampere force I * B * L.

The minimum value of the induction vector ensures uniform motion. We equate the two forces and we get equality:

k * m * g = I * B * L;

Let us express the induction:

B = (k * m * g) / (I * L) = (0.6 * 0.14 * 10) / (50 * 0.3) = 0.056 T.

The direction of the current is directed along the bar perpendicular to the rails.