Question

A conductor bent into a semicircle of total length 3.88 m is placed in a 2.0 T magnetic field perpendicular to the plane of the semicircle. What is the magnitude of the total force on the conductor if a 5.02 A current passes through it?

Answer 1

Answer:

Magnetic force, F = 38.95 N

Explanation:

Given that,

Length of the conductor, L = 3.88 m

Magnetic field, B = 2 T

Current flowing in the conductor, I = 5.02 A

The magnetic field perpendicular to the plane of the semicircle. The angle between the magnetic field and the plane is 90 degrees. The expression for the magnetic force is given by:

$F=ILB\ sin\theta$

$F=ILB$

$F=5.02\times 3.88\times 2$

F = 38.95 N

So, the magnitude of the total force acting on the conductor in the magnetic field is 38.95 N. Hence, this is the required solution.