Magnetic flux outside cylindrical conductor

In summary, the conversation discusses the problem of showing that the magnetic flux outside a cylindrical conductor is zero. The problem involves a coaxial cable with distributed currents flowing in opposite directions in the two conductors. Expressions for the magnetic flux density in three different regions are derived using Ampere's law. The conversation also addresses the issue of the magnetic flux being zero in the region outside the cable and provides a solution using a closed path and the enclosed current. The summary concludes with a mention of an error in the third expression for the magnetic flux density and a correction is suggested.
  • #1
mmh37
59
0
Hey everyone,

I have failed to show that the magnetic flux outside a cylindrical conductor is zero.

The problem goes like this:

a coaxial cable consists of a solid inner cylindrical conductor of radius a and an outer cylindrical conductor of inner and outer radius b and c. Distributed currents of equal magnitude I flow in opposite directions in the two conductors. Derive expressions for the magnetic flux density B(r) for each of these regions

1) 0<r <a

by amperes law I showed: [tex] B = \frac {K *l*r} {2*pi*a^2} [/tex]

where K is the permeability of free space, don't know how to write that with latex

2) a<r<b

again by amperes law: [tex] B = \frac {K *l} {2*pi*r} [/tex]

3) b<r<c

again by amperes law: [tex] B = \frac {K *l*(r^2-b^2)} {2*pi*r*(c^2-b^2)} [/tex]

4) c<r

I got stuck here, surely it must be zero...but how can that be shown?

thanks for your help - it's very much appreciated
 
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  • #2
With the coaxial cable coming out of the paper at you, draw a circular closed path around the outside of it.

INTEG[B dot dL] around that closed path = mu * I(enclosed)

The total I enclosed by the circle is zero, since the current up the inner conductor is the same as the current down the outer conductor, so B (or H) has to be identically zero everywhere outside the coax.


Edit -- BTW, check your answer for (3). When r=c, you should get zero for H.
 
Last edited:
  • #3
Thanks Berkeman - that's explained it and it's very much appreciated! :-) As for the third one it surely has to be (c^2-r^2)/(c^2-b^2)
 

1. What is magnetic flux outside a cylindrical conductor?

Magnetic flux outside a cylindrical conductor refers to the magnetic field lines that exist outside the surface of the conductor. These field lines are created by the flow of electric current through the conductor and can be measured in units of Weber (Wb) or Tesla (T).

2. How is the magnetic flux outside a cylindrical conductor calculated?

The magnetic flux outside a cylindrical conductor can be calculated using the formula Φ = B x A, where Φ is the magnetic flux, B is the magnetic field strength, and A is the area perpendicular to the magnetic field. Alternatively, it can also be calculated using the formula Φ = μ x H x A, where μ is the permeability of the medium and H is the magnetic field intensity.

3. What factors affect the magnetic flux outside a cylindrical conductor?

The magnetic flux outside a cylindrical conductor is affected by several factors, including the strength of the electric current flowing through the conductor, the distance from the conductor, the permeability of the medium, and the shape and size of the conductor.

4. How does the magnetic flux outside a cylindrical conductor change with distance?

The magnetic flux outside a cylindrical conductor decreases as the distance from the conductor increases. This is because the magnetic field strength decreases with distance, leading to a decrease in the number of magnetic field lines passing through a given area.

5. How can the magnetic flux outside a cylindrical conductor be increased?

The magnetic flux outside a cylindrical conductor can be increased by increasing the strength of the electric current flowing through the conductor, using a higher permeability medium, or increasing the area perpendicular to the magnetic field. Additionally, wrapping the conductor in a coil or adding a ferromagnetic material can also increase the magnetic flux.

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