Register to reply

Magnetic field in coaxial cables

by A_B
Tags: cables, coaxial, field, magnetic
Share this thread:
May5-11, 03:03 AM
P: 85
I think I solved this problem correctly, I just want to check because it's an interesting problem, and it's from a past exam.

1. The problem statement, all variables and given/known data
A coaxial cable consists of an inner solid conductor of radius [itex]r_1[/itex], and and outer conducting cylindrical shell of inner radius [itex]r_2[/itex] and outer radius [itex]r_3.[/itex] In both conductors, the current equals [itex]I_0[/itex], but they are in opposite directions. Also, the current distribution is not uniform, it increases linearly with distance from the central axis. In the inner conductor is varies as [itex]j_1=C_1 r[/itex] and in the outer cable as [itex]j_2=C_2 r[/itex].

Calculate the magnetic induction (which I take means magnetic field strength) as function of [itex]R[/itex], the distance from the central axis, and [itex]I_0[/itex] for:

a) [itex]R < r_1[/itex]
b) [itex]r_1 < R < r_2[/itex]
c) [itex]r_2 < R < r_3[/itex]
d) [itex]R > r_3[/itex]

2. Relevant equations
I think the static Ampère's law is the correct equation
[tex]\oint \bar{B} \cdot d\bar{s} = \mu_0 I[/tex]
Where I is the current through any surface which has the curve of the LHS as boundary.

3. The attempt at a solution

a) We consider a circle of radius [itex]R < r_1[/itex] centered at the central axis of the coaxial cable in the plane perpendicular to the current. The LHS of Amp's law for this circle gives [tex]\oint \bar{B} \cdot d\bar{s} = 2\pi B R.[/tex]
For the RHS we need to find the total current through the circle.
I &= \iint\limits_{circle} j_1 dA
\\ &= \int_0^{2\pi} d\theta \int_0^R C_1 r^2 dr
\\ &= \frac{2}{3} \pi C_1 R^3

If we take this integral from [itex]0[/itex] to [itex]r_1[/itex] , we know that the current through the circle is [itex]I_0[/itex]. Thus, we can figure out the constant [itex]C_1.[/itex]
[tex] C_1 = \frac{3I_0}{2\pi r_1^3} [/tex]
We put this in the previous formula for the current to obtain:
[tex] I = \left(\frac{R}{r_1}\right)^3 I_0 [/tex]

We plug it all into Ampère's law to obtain
[tex]B=\frac{\mu_0 R^2}{2\pi r_1^3}I_0[/tex]

b) When [itex]r_1 < R < r_2[/itex] the current through the circle is always [itex]I_0[/itex] and the countour integral still equals [itex]2\pi BR[/itex] so the magnetic field strength is
[tex] B = \frac{\mu_0 I_0}{2\pi R}[/tex]

c)The current through the circle is [itex]I_0[/itex] minus the current through the bit of the outer conductor that goes through the circle. We shall first calculate the current through this bit.
I_2 &= \iint\limits_{r_2 \rightarrow R} j_2 dA
\\ &= \int_0^{2\pi} d\theta \int_{r_2}^{R} C_2 r^2 dr
\\ &= \frac{2}{3} \pi C_2 \left(R^2 - r_2^3\right)

Taking this integral to [itex]r_3[/itex] should give [itex]I_0[/itex] so we can figure out the constant [itex]C_2[/itex].
[tex]C_2 = \frac{3I_0}{2\pi \left(r_3^3-r_2^3\right)}[/tex]

Putting it all together gives a total current through the circle of:
[tex]I = I_0 - I_2 = I_0 - \frac{\left(R^3 - r_2^3\right) I_0}{\left(r_3^3 - r_2^3\right)}[/tex]

Applying Amp's law gives
[tex] B = \frac{\mu_0 I_0}{2 \pi R} \left( 1- \frac{R^3 - r_2^3}{r_3^3 - r_2^3} \right)[/tex]

c) The total current is zero, therefore, so is the magnetic field.
[tex] B = 0[/tex]
Phys.Org News Partner Science news on
Wildfires and other burns play bigger role in climate change, professor finds
SR Labs research to expose BadUSB next week in Vegas
New study advances 'DNA revolution,' tells butterflies' evolutionary history

Register to reply

Related Discussions
The Magnetic Field of a Coaxial Cable Introductory Physics Homework 0
The Magnetic Field of Coaxial Cables Advanced Physics Homework 1
Magnetic and electric field between current carrying coaxial cables Introductory Physics Homework 1
Coaxial cable and magnetic field Introductory Physics Homework 8
Magnetic field of a coaxial cable Introductory Physics Homework 1