Asingle-turn souare loop of wire 1.000m on a side arries solenoid has 30 turns p

Question

Asingle-turn souare loop of wire 1.000m on a side arries solenoid has 30 turns per centimeter and carries a counterclockwise arrett of 0.300 ?. The loop S inside a solenoid, with the plane of th-loop pendcular to the magnetic held of the solond The current of 20.0 A Find the force on each side of the loop and the torque acting on the loop response is within 10% of the correct value This may be dae to rondiff error or you could have místake in your allation. TY ??t alintermedal, results to at lent leur digt norxy to minut roundolt ertor N (top side) The correct anower s not rero N (bottom side) N left side) N (right side) N-m (torque) Need Hel

Explanation / Answer

Write the exprerssion for the magnetic field in the solenoid -  

B = (mu0)*(N / L )*I

I = current in the solenoid

N/L = 30*100

mu0 = 4*pi*10^-7

So :

B = 4*pi*10^-7*30*100*10 = 12*pi*10^-3 T

Now, we have the magnetic field that is acting on the square loop of wire.

So, the force on each side of the loop will be :

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

B = 12*pi*10^-3

I = 0.3 A

L = (1/100) meters

the angle is 90 degrees, because as metioned in the problem, that the plane fo the loop is perpendicular to the magnetic field

so : F = 0.3*(1/100)*12*pi*10^-3 = 36*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 = 36*pi*10^-6 = 1.13 x 10^-4 N

Again, for determining the torque -  

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

B = 12*pi*10^-3 (Tesla)

sin(90) = 1,

So, the torque acting on the loop = 0.3*(1/100)^2*12*pi*10^-3*1 = 3.6 x 10^-7 N.m

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