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 0.250 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. magnitude 282500 What is the direction of the magnetic field inside a solenoid. mu N direction directed away from the center Find the magnitude of the torque acting on the loop. 0 N middot m

Explanation / Answer

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

So : B = 4*pi*10^-7*30*100*15 = 0.05654 (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 = 0.05654
I = 0.250 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 = 0.250x(2/100)x0.05654 = 2.827x10^-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 = 2.827x10^-4 (Newtons)

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