A wire bearing current of 1.25 A crosses a region in which there is a uniform B field in the +y direction.The wire is inclined at an angle of 75 degrees above the +x direction.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 = I L B sin(theta), where I is the current, L is the length of the wire in the magnetic field, B is the strength of the magnetic field, and theta is the angle between the wire and the magnetic field. In this case, the current is 1.25 A, the length of the wire in the magnetic field is 15 cm, the magnetic field strength is 0.5 T, and the angle between the wire and the magnetic field is 75 degrees. Plugging these values into the equation, we get F = (1.25 A) (0.15 m) (0.5 T) sin(75 degrees) = 0.19 N.

A Wire Bearing Current of 1