A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise cur

Question

A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A.

(a) Find the force on each side of the loop.

(b) Find the magnitude of the torque acting on the loop.
_____ N · m

Explanation / Answer

solution:

The magnetic field in the solenoid is :

B = Uo*(N / L )*I

I = current in the solenoid

N/L = 30*100

Uo = 4*pi*10^-7 constant

So :

B = 4*pi*10^-7*30*100*15 = .0565 (Tesla)

We have the magnetic field that is acting on the square loop of wire.

Now, we know that the force on each side of the loop will be :

F = I*L*B*SIN(angle)

B = .0565 T

I = 0.240 Ampere

L = (2/100) =.02 meters

the angle is 90 degrees, because you said, that the plane fo the loop is perpendicular to the magnetic field

so : F = 0.24*(.02)*.0565 = 2.71e-4 (Newtons)

This force will have the same value for each side, because it's a square, so the have the same lenght, and also, because the magnetic field is perpendicular to each side.

F1 = F2 = F3 = F4 = 2.71e-4 (Newtons)

Now, the torque, here you need to know :

Torque = N*I*A*B*sin(angle)

B = .0565(Tesla)

sin(90) = 1, the torque acting on the loop :

N =1

1*0.24*(.02)^2*.0565 = 5.426e-6 N.m ans

Hope that might help you

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