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Jackson 5 6 Homework Solution

The document summarizes the solution to a homework problem involving using Ampere's law and superposition to find the magnetic field inside a hole drilled in a cylindrical conductor. The conductor has a uniform current density J0 parallel to its axis. By modeling the current distributions with and without the hole separately, and then combining the results, the magnetic field inside the hole is found to be directed along the axis of the cylinder (j-direction) with a magnitude of μ0J0d/2, where d is the distance of the hole from the cylinder axis.

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Faheem Afsar
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651 views2 pages

Jackson 5 6 Homework Solution

The document summarizes the solution to a homework problem involving using Ampere's law and superposition to find the magnetic field inside a hole drilled in a cylindrical conductor. The conductor has a uniform current density J0 parallel to its axis. By modeling the current distributions with and without the hole separately, and then combining the results, the magnetic field inside the hole is found to be directed along the axis of the cylinder (j-direction) with a magnitude of μ0J0d/2, where d is the distance of the hole from the cylinder axis.

Uploaded by

Faheem Afsar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Jackson 5.

6 Homework Problem Solution


Dr. Christopher S. Baird
University of Massachusetts Lowell

PROBLEM:
A cylindrical conductor of radius a has a hole of radius b bored parallel to, and centered a distance d
from, the cylinder axis (d + b < a). The current density is uniform throughout the remaining metal of
the cylinder and is parallel to the axis. Use Ampere's law and principle of linear superposition to find
the magnitude and direction of the magnetic-flux density in the hole.

SOLUTION:

a a
d b d b
= +

Define the direction in which the hole is displaced from the cylinder axis as the positive x direction,
and the direction coming out of the page as the positive z direction. If a uniform current density J0
flows in the positive z direction everywhere colored blue in the leftmost image, this is equivalent to a
uniform current density J0 flowing in the positive z direction everywhere in the cylinder plus a uniform
current density J0 flowing in the negative z direction in the hole region.

Ampere's Law in integral form states:

∮C B⋅d l=0 ∫S J⋅n da


Draw a circular Amperian at some radius r in side the cylinder with no hole and take the line integral
over this loop. Because of the symmetry, the magnetic field will be constant along this integral and can
come out of the integral.

B∮C d l= 0 J 0∫S da

B 2  r= 0 J 0  r 2

1 
B=  0 J 0 r 
2

1
B=  0 J 0 − y i x j
2
Using the same method, place the hole-shaped region with current flowing into the page at the origin
and it produces the magnetic field:

1
B=−  0 J 0 −y ix j
2

Now shift this region to its proper place, x → x - d

1
B=−  0 J 0 −y i x−d  j
2

And add together the two field to get the final answer:

1 1
B=  0 J 0 − y i x j−  0 J 0 − y i x−d  j
2 2

d
B=  0 J 0 j
2

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