A long wire carrying 450 A of current makes two 90 degree be

A long wire carrying 4.50 A of current makes two 90 degree bends as shown below. The bent part of the wire passes through a uniform 0.24 T magnetic field directed as shown. The magnetic field is limited to a space of 30 cm by 30 cm. Find the magnitude of the force that the magnetic field exerts on the curve.

Solution

Here ,

current , I = 4.5 A

B = 0.24 T

force on horizontal part of wire

F1 = B*I*0.30

F1 = 4.5 * 0.24 * 0.30 = 0.324 N

force on the vertical part of the wire

F2 = B*I*0.10

F2 = 4.5 * 0.24 * 0.10 = 0.108 N

magnitude of net force = sqrt(0.324^2 + 0.108^2)

magnitude of net force = 0.341 N

the magnitude of net force acting on the wire is 0.341 N