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 = 2.92 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

At point a, the distance is r = 0.001 m:

B = µI1 / 2r = ((4 * 10^-7 Tm/A) * 1.14 A) / (2 * 0.001 m)

= 2.28 x 10^-4 T; Direction is upwards (towards the top)



At point b, the distance is R = 0.003 m :

At this point, we need to account for both effects from current I1 and I2.

. Effect by current I1 onto point b:

B1 = µI1 / 2R = ((4 x 10^-7 Tm/A) x 1.14 A) / (2 x 0.003 m)

= 0.76 x 10^-4 T; Direction is upwards (towards the top)


. Effect by current I2 onto point b:

B2 = µI2 / 2R = ((4 x 10^-7 Tm/A) x 2.92 A) / (2 x 0.003 m)

= 1.94 x 10^-4 T; Direction is downwards (towards the bottom).

Therefore, B at point b will be B = B1 + B2 = (1.94+0.76)*10^-4 = 2.7*10^-7 T

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