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 = 1.06 A out of the page and the current in the outer conductor is I2 = 3.16 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

The system has axial symmetry and therefore the electric field can be found from the Ampere-Maxwell equation. Considering closed loops, concentric with the cable and passing by the points of interest, the line integral can be related to the radius of each loop and the current flowing through the surface spanned by the loop.

a) At point A , B=2piRa=Bds cos0=integration B.ds=0 Ia

B= 0 Ia/2piRa=4pi*10^-7*1.06/2pi*(0.001)= 2.12e-4 T in the “positive” y-direction

b) At point B,   B=2piRb=Bds cos0=integration B.ds=0 (-Ib)

B= -0 Ib/2piRb=-4pi*10^-7*(-3.16)/2pi*(0.003)= 2.10e-4 T in the “negative” y-direction

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