A cross-sectional view of a coaxial cable. The center conductor is surrounded by

Question

A cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is I_1 = 1.12 A out of the monitor, and the current in the outer conductor is I_2 = 2.88 A into the monitor. Determine the magnitude and direction of the magnetic field at point a. Determine the magnitude and direction of the magnetic field at point b.

Explanation / Answer

a)

At point a, the distance is r = 0.001 m. At the same point...

B = µI1 / 2r = ((4 x 10^-7 Tm/A) x 1.12 A) / (2 x 0.001 m)

= 2.24 x 10^-4 T;

Direction is upwards (towards the top).

b)

At point b, the distance is R = 0.003 m. At the same point...

B = µI2 / 2R = ((4 x 10^-7 Tm/A) x 2.88 A) / (2 x 0.003 m)

=1.92 x 10^-4 T;

Direction is downwards (towards the bottom).

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