Question

An 8.00-cm-long piece of wire is formed into a square, and carries a clockwise current of 0.150 A. The loop is placed inside a solenoid, and the plane of the loop is perpendicular to the solenoid’s magnetic field. The solenoid carries a counterclockwise current of 17.0 A, and has 25 turns per centimeter. What is the force on each side of the loop?

Please explain

Answer 1

The torque on a current carrying loop on a magnetic field is given by

I is the current flowing through the loop.

Here, I = 0.150 A

A is the area of the loop.

Here, the length of the wire used to make the loop = 8 cm = 0.08 m

So, side of the square loop is

a = 0.08/4 = 0.02 m

So, area of the loop A = 0.02*0.02 = 0.0004 m²

B is the magnetic field

The magnetic field due to a solenoid is given by

Here, N = 25 turns per cm = 2500 turns per meter.

I = 17.0 A

So,

No, it is given that the loop and the magnetic field are perpendicular to each other.

So,

Now, consider the view of the solenoid as shown below.

The force will be in the direction of

The magnitude of the force is given by

the magnetic field is in the +z direction.

For the top side of the wire, the direction of current is in the +x direction

So, Force is in xXz = -y direction

For the bottom side of the wire, the direction of current is in the -x direction

So, Force is in -xXz = +y direction

For the left side of the wire, the direction of current is in the +y direction (upward)

So, Force is in yXz = x direction

For the right side of the wire, the direction of current is in the -y direction

So, Force is in -yXz = -x direction