Question

ure below is a view of a coaxial cable. In a particular application, the in the front 6. The Fig current in the inner conductor is 4.00 A to the right (out of the paper view) and the current fonr in theouter conductor is 1.00 A to the left (into the paper in the view). The radius of the inner conductor is 2.0 cm, the inner radius of the outer Law conductor is 4.0 cm and the outer radius is 5.0 cm. Applying Ampere's determine magnitude and direction of the magnetic field: teco of the magnetic fiel (a) 0.5 cm from the center of both conductor b) 3.0 cm from the center (c) 6.0 cm from the center (d) if the intensity of the current in both conductors is the same.

Answer 1

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I_1 = 4 A I_2 = 1 A r_1 = 2 cm r_2 = 4 cm r_3 = 5 cm Ampere's law implies integral B^vector middot dl^vector = mu_0 I_en a) For r = 0.5 from center of the loop I_en = I/pi r_1^2 pi r^2 = 4 times (0.5)^2/4 = 0.25 A B middot 2 pi r = mu_0 times 0.25 B = mu_0 times 0.25 = 4 pi times 10^-7 times 0.25/2 pi times 0.5 times 10^-2 = 10^-5 T B = 1.0 times 10^-5 T direction implies anticlockwise in front view b) For r = 3 cm from water I_en = 4 A B middot 2 pi r = mu_0 times 4 implies B = 4 pi times 10^-7 times 4/2 pi times 3 times 10^-2 = 2.67 times 10^-5 T B = 2.67 times 10^-5 T direction implies anticlockwise (front view) c) For r = 6 cm from center I_en = (4 - 1) = 3