The figure below is a cross-sectional view of a coaxial cable. The center conduc

Question

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.

Explanation / Answer

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)

Let me know if you have any doubt.

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