Find the force on each side of the loop

[Roze]

Homework Statement

A single-turn square loop of wire 2.00 cm on a side carries a counterclockwise current of .200A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magentic field of the solenoid. The solenoid has 30 turns per centimeter and carries a counterclockwise current of 15.0A. Find the force on each side of the loop and the torque acting on the loop.

Homework Equations

B=unI

The Attempt at a Solution

I'm really confused about visualizing this situation, and figuring out what exactly their asking for. I guess that's my first problem.

 

[Shooting Star]

Hint: The magnetic field inside a solenoid is uniform. Just treat it as a region of space where B is constant.

 

[Moetasim]

The force on each side of the loop can be found by using the equation F = IL x B, where I is the current, L is the length of the side of the loop, and B is the magnetic field. In this case, the magnetic field is created by the solenoid and can be calculated using the equation B = u*n*I, where u is the permeability of the material (usually air or vacuum), n is the number of turns per unit length, and I is the current in the solenoid.

To find the force on each side of the loop, we need to first calculate the magnetic field at the location of the loop. Since the loop is inside the solenoid, the magnetic field will be uniform and perpendicular to the plane of the loop. Therefore, we can use the equation B = u*n*I to calculate the magnetic field at the center of the loop.

B = 4*pi*10^-7 * (30 turns/cm) * (15A) = 0.018 T

Now, we can use the equation F = IL x B to calculate the force on each side of the loop. Since the loop has four sides, we need to calculate the force on each side separately.

On the top and bottom sides of the loop, the force will be in the same direction and have the same magnitude. The length of these sides is 2 cm, and the current is 0.2A. Therefore, the force on each side is:

F = (0.2A)(2 cm)(0.018 T) = 0.0072 N

On the left and right sides of the loop, the force will be in the opposite direction and have the same magnitude. The length of these sides is also 2 cm, and the current is still 0.2A. Therefore, the force on each side is:

F = (0.2A)(2 cm)(0.018 T) = 0.0072 N

The total force on the loop is the vector sum of these four forces. Since they are all equal in magnitude and direction, the total force on the loop is:

F = 4(0.0072 N) = 0.0288 N

To find the torque acting on the loop, we can use the equation torque = N x B, where N is the number of turns in the loop and B is the magnetic field. Since the loop