A Calculate the y component of the magnetic field at the cen

A) Calculate the y component of the magnetic field at the center of the square.

B) Calculate the x-component of the force on a 1.0-cm long piece of the lower right-hand wire, due to the other three wires.

The four wires that lie at the corners of a square of side a= 3.00 cm are carrying equal currents i= 2.60 A into (+) or out of (-) the page, as shown in the picture.

Solution

According to the given problem,

Use the force eaution for a magnetic field,

F = I*L*B

First calculate,B magnetic field on the lower right wire

B = ₀*I/2r but ₀/2 = 2*10^-7)

Now the lower - wire produces a field that is down and with magnitude,

B = 2*10^-7*2.60A/0.03

B = 1.733 10^-5 T

The +ve wire produces the same field but is directed to the right

And the other -ve wire has a distance = 0.030*cos45° = 0.0212m

Magnetic field at this point is,

B = 2*10^-7*2.60/0.0212

B = 2.451 *10^-5T but this is directed down and to the left 45°

A) Now the components are

B_x = 1.733 10^-5 - 2.451 *10^-5*cos45° =

B_x = 0 T

Then y-component is down

B_y = 1.733 10^-5 - 2.451 *10^-5*cos45°

B_y = 3.467 *10^-5 T [A]

B) Since we want the x component of the force then we only need the B_y

As I is out and B is down the resulting force will be in the +x direction

F = I*L*B =2.60*0.01*3.467 *10^-5

F = 9.013 *10^-7 T [B]