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 t_1 = 1.16 A out of the page and the current in the outer conductor is I_2 = 3.18 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. magnitude mu T direction upward (b) Determine the magnitude and direction of the magnetic field at point b. magnitude mu T direction downward

Explanation / Answer

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

B = µI1 / 2r = ((4 x 10^-7 Tm/A) x 1.16 A) / (2 x 0.001 m)

= 232 x 10^-6 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. So at point b, there will be two calculation sets:

. Effect by current I1 onto point b:

B1 = µI1 / 2R = ((4 x 10^-7 Tm/A) x 1.16 A) / (2 x 0.003 m)

= 77.3 x 10^-6 T; Direction is upwards (towards the top)


Effect by current I2 onto point b:

B2 = µI2 / 2R = ((4 x 10^-7 Tm/A) x 3.18 A) / (2 x 0.003 m)

= 212 x 10^-6 T; Direction is downwards (towards the bottom).

Therefore, B at point b will be B = B1 + B2 = 289.3 × 10^-6 T
NOTE: watch-out for the sign +/- since one has direction up and the other has direction down

In general, direction will be dominated by direction of B2 [effected by the current I2] since the point b is closer to I2 and moreover, value of B2 is much greater than B1.

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