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 = 1.04 A out of the page and the current in the outer conductor is I2 = 3.02 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

a) by Ampere's law

integral (B*dl) = mu_o*I_enclosed

B*2*pi*d = mu_0*I1

B = mu_o*I1/(2*pi*d)

B = (2*10^-7*1.04)/(1*10^-3) = 208*10^-6 T = 208 uT
direction upward

b) by ampere's law

B = mu_o*(I1-I2)/(2*pi*3d)

B = (2*10^-7*(1.04-3.02))/(3*1*10^-3)

B = -132 uT

direction downward

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