Question

The drawing shows four insulated wires overlapping one another, forming a square with 0.050-m sides. All four wires are much longer than the sides of the square. The net magnetic field at the center of the square is 77 ?T directed into the drawing. Calculate the current I.
3.7 A 7.5 A 4.8 A

Answer 1

field at the center due to a finite wire

B = (u0*I/4*pi*r)*(cos45-cos135) = u0*I*sqrt2/4*pi*r

r = L/2 = 0.05/2 = 0.025 m

i1 = 3.7

i2 = 7.5 A

i3 = 4.8 A

i4 = I

B1 = -(uo*I1*sqrt2/(4*pi*r)) = - (4*pi*10^-7*3.7*sqrt2)/(4*pi*0.025) = -21 uT   (into)

B2 = -(uo*I2*sqrt2/(4*pi*r)) = - (4*pi*10^-7*7.5*sqrt2)/(4*pi*0.025) = -42.4 uT (into)

B3 = (uo*I3*sqrt2/(4*pi*r)) = -(4*pi*10^-7*4.8*sqrt2)/(4*pi*0.025) = -27.1 uT   (into)

B4 = (uo*I1*sqrt2/(4*pi*r)) = (4*pi*10^-7*I*sqrt2)/(4*pi*0.025) = 21 uT   (out)

Bnet = B1 + B2 + B3 + B4

-77 = -21 - 42.4 - 27.1 + 5.65I

I = 2.4 A <<<<<---------------answer