A circular loop of current-carrying wire, diameter d1 sits to the left of a rect

Question

A circular loop of current-carrying wire, diameter d1 sits to the left of a rectangular loop of currentcarrying wire of width d3. Both loops lie in the same plane. The current, I1 in the circular loop is in a counterclockwise direction. If the edges of the loops are d2 apart, and the magnetic field at the center of the circular loop is zero, what is the magnitude and direction of the current in the rectangular loop? Answer in terms in I1, d1, d2, d3, and appropriate constants. And, yes, it’s a messy looking, not particularly obvious thing. (The long sides of the rectangular are suitably long to use a long-wire approximation.)

Explanation / Answer

Magnetic filed at the center of the circular loop is B1 = mu_0*i/(2*R)

i is the current = i1

R is the radius of the loop = d1/2

then B1 = mu_0*i1/(2*d1/2) = mu_0*i1/d1.......(1)
to get the magnetic field zero at the center of the circular loop

the current in the rectangular loop must be in the counter clockwise

due to near side of the rectangular loop the magnetic field at the cednter of the circular loop is B2 = mu_0*i2/(2*pi*(d2+0.5d1))

i2 is the required current

the direction of magnetic field B2 is into the plane

due to far side of the rectangular loop the magnetic field at the cednter of the circular loop is B3 = mu_0*i2/(2*pi*(0.5d1+d2+d3))

the direction of B3 is out of the page

then Net magnetic filed at the center of the circular loop is

B1+B2+B3 = 0

[mu_0*i1/(d1)]-[mu_o*i2/(2*pi*(d2+0.5*d1))]+[mu_o*i2/(2*pi*(0.5d1+d2+d3))] = 0

B2 is taken negative because the direction is into the plane where as B1 and B3 are out of the plane

(i1/d1) - (i2/(2*pi))[(1/(0.5*d1+d2)) + (1/(0.5d1+d2+d3))]

i1/d1 = (i2/(2*pi))[(1/(0.5*d1+d2)) + (1/(0.5d1+d2+d3))]

then i2 = (2*pi*i1/d1)/ [(1/(0.5*d1+d2)) + (1/(0.5d1+d2+d3))]

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