A coaxial rod is composed of an inner solid conductor of radius 2.54 cm that is

Question

A coaxial rod is composed of an inner solid conductor of radius 2.54 cm that is surrrounded by an insulating shell that expands to radius 3.49 cm, which is itself surrounded by a solid conducting shell that extends to radius 4.45 cm. A current of 122 A is carried into the page by the inner conducter and back out of the page by the outer conducting shell. The current density WITHIN EACH conductor is UNIFORM though they may have DIFFERENT current densities. Determine the following using Amper's LAw:

a)The magnitiude of the magnetic field at r = 1.18 cm (within the inner conducter)

b)The magnitude of the magnetic field at r = 3.05 cm (within the insulating shell)

c)The magnitude of the magnetic field at r = 3.87 cm (within the conducting shell)

d)The magnitude of the magnetic field at r = 5.12 cm (outside the rod)

(Please show all the steps and explanations. Thanks!)

Explanation / Answer

1. a.

ampere's law

B 2 pi r = u0 I enc

so Ienc = I Aenc/A = I pi r^2/ pi R^2

= I r^2/R^2

B 2 pi r = u0 I r^2/R^2

B =2*10^(-7)*122*1.18E-2/(2.54E-2)^2 = 4.46E-4 T

b. now contains all of inner I, so Ienc = 122

B = 2*10^(-7)*122/(3.05E-2)= 8.0E-4 T

c)

now contains some of outer charge

I outer enc = I*Aenc/A = -122 (r^2 - 3.49E-2^2)/(4.45E-2^2-3.49E-2^2)

Itotal = 122 -122 (r^2 - 3.49E-2^2)/(4.45E-2^2-3.49E-2^2)

so

B*2*pi*3.87E-2 = 4*pi*10^(-7)*(122 -122 (3.87E-2^2 - 3.49E-2^2)/(4.45E-2^2-3.49E-2^2))

solve for B
B=3.99E-4 T

d) now contains all of outer I

so I enc = 122 - 122 = 0

so B = 0

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