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 0.170 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

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 = 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 = 0.17 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 = 0.17*(2/100)*18*pi*10^-3 = 61.2*pi*10^-6 (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 = 61.2*pi*10^-6 (Newtons)

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 :

0.17*(2/100)^2*18*pi*10^-3*1 = 122.4*10^-8 N.m

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