An electromagnet produces a magnetic eld of 0.95 T in a cylindrical region of radius 3.0 cm between its poles. A straight wire carrying a current of 15.0 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic eld. What magnitude of force is exerted on the wire?

0.855 N

Explanation:

Using

F = BILsinФ ... Equation 1

Where F = magnitude of the force exerted on the wire, B = magnetic Field, I = current, L = Length of the wire in the magnetic Field, Ф = angle between the wire and the magnetic field.

Given: B = 0.95 T, I = 15.0 A, L = 2r, r = radius = 3.0 cm = 0.03 m L = 2*0.03 = 0.06 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.95 (15) (0.06) sin90°

F = 0.855 N.

Hence the magnitude of force that is exerted on the wire = 0.855 N