A conductor with a current of 5 A is placed in a uniform magnetic with an induction of 1 x 10 -2 T

A conductor with a current of 5 A is placed in a uniform magnetic with an induction of 1 x 10 -2 T. The angle between the directions of the current and the field is 60 degrees. Determine the active part of the conductor if the field acts on it with a force of 2 x 10 -2 N.

Given:

I = 5 A;

B = 1 * 10-2 T;

α = 60 °;

F = 2 * 10-2 H.

To find:

l -?

Decision.

FA = B * I * l * sin α;

2 * 10-2 = 1 * 10-2 * 5 * l * sin 60 °;

l = 2 * 10-2 / (1 * 10-2 * 5 * sin 60 °);

l = 2 * 10-2 / (10-2 * 5 * √3 / 2);

l = 2 / 5√3 / 2;

l = 4 / 5√3 = 4/5 * 1.7 = 4 / 8.5 = 0.5 m.

Answer: The active portion of the conductor is approximately 0.5 meters.