The figure below is a crosssectional view of a coaxial cable

The figure below is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular application, the current in the inner conductor is I₁ = 1.14 A out of the page and the current in the outer conductor is I₂ = 3.14 A into the page. Assuming the distance d = 1.00 mm,answer the following.

(a) Determine the magnitude and direction of the magnetic field at point a.

(b) Determine the magnitude and direction of the magnetic field at point b.

Solution

1. At point A

B*2*pi*rA = integral(B*ds*cos 0) = integral(B.ds)

B*2*pi*rA = uo*Ia

B = u0*Ia/(2*pi*rA)

B = 4*3.14*10^-7*1.14/(2*3.14*10^-3) = 2.28*10^-4 T (in the upward direction)

2.

At point B

B*2*pi*rB = integral(B*ds*cos 0) = integral(B.ds)

B*2*pi*rB = uo*(Ia - Ib)

B = -u0*(Ia - Ib)/(2*pi*rB)

B = 4*3.14*10^-7*2.14/(2*3.14*3*10^-3) = -1.43*10^-4 T (in the downward direction)

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