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

Question

single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.230 A. The loop is inside a solenoid, with the plane of the loop perpendicular to he 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. 1.30 magnitude What is the direction of the magnetic field inside a solenoid. HN direction directed away from the center v (b) Find the magnitude of the torque acting on the loop

Explanation / Answer

The magnetic field in the solenoid is :

B = Uo*(N / L )*I

I = current in the solenoid = 15 A

N/L = 30*100
Uo = 4*pi*10^-7

So :
B = 4*pi*10^-7*30*100*15 = 18*pi*10^-3 (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 = 18*pi*10^-3
I = 0.23 Ampere
L = (2/100) meters

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

so : F = 0.23*(2/100)*18*pi*10^-3 = 8.28*pi*10^-5 (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 = 8.28*pi*10^-5 (Newtons) = 2.6 x 10-4 N

Now, the torque, here you need to know :

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

B = 18*pi*10^-3 (Tesla)

sin(90) = 1, the torque acting on the loop :
torque = 30*0.23*(2/100)2*18*pi*10^-3*1 = 1.6*10-4 N.m

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