A wire bearing current of 1.25 A crosses a region in which there is a uniform B field of magnitude 1.0 T in the +y direction. The wire is inclined at an angle of 75 degrees above the +x direction, in the xy-plane. What is the magnitude of the force on the wire, if the magnetic field region spans 15 cm in the x direction?
A) 0.19 N
B) 0.70 N
C) 0.18 N
D) 0.05 N

Question

Quizplus · Accepted Answer

The force on a current-carrying wire in a magnetic field is given by the equation F = BIL sin(theta), where B is the magnetic field strength, I is the current, L is the length of the wire in the magnetic field, and theta is the angle between the wire and the magnetic field. In this case, B = 1.0 T, I = 1.25 A, L = 15 cm = 0.15 m, and theta = 75 degrees. Plugging these values into the equation, we get F = (1.0 T)(1.25 A)(0.15 m)sin(75 degrees) = 0.19 N.

A Wire Bearing Current of 1