How Does an Unbalanced Coaxial Cable Affect Enclosed Current Calculations?

[satchmo05]

Homework Statement

I just have a question concerning an unbalanced coaxial cable. By unbalanced, I mean that the inner and outer currents are not equal magnitudes of the opposite direction of each other. My homework problem has an inner current of +100[mA] and an outer current of 90[mA] in the opposite direction. My question is regarding the form of the answer.

Homework Equations

Cylindrically symmetric current distribution - able to use Ampere's Law here!
∫B•dl = μ_o*I_enc

The Attempt at a Solution

Implementing the cylindrically symmetric current distribution, the left side of the equation ends up being:
2∏ρ*B_Ф(ρ) = ...
However, the right side of the equation is where I am having trouble. Would the enclosed current be equal to +10[mA] (100-90[mA]), or something else? An enclosed current of 10[mA] to me makes sense, because if I were an observer looking at the cable, I would only see a magnetic field rotating in the phi-direction, with a current of 10[mA] flowing up the z-axis. I appreciate all help in advance! Thank you much!

 

[anathema]

Thank you for your question about an unbalanced coaxial cable. In order to determine the form of the answer, we need to consider the principles of Ampere's Law and the concept of enclosed current.

As you correctly noted, the left side of the equation for Ampere's Law in a cylindrical coordinate system is 2∏ρ*BФ(ρ). This represents the magnetic field along a closed path around the cable, with the path being at a distance ρ from the axis of the cable.

On the right side of the equation, we have the enclosed current, which is the sum of all currents passing through the closed path. In the case of an unbalanced coaxial cable, there are two currents - one in the inner conductor and one in the outer conductor. These currents are in opposite directions, so we can subtract them to find the net enclosed current.

In your example, the inner current is +100[mA] and the outer current is -90[mA]. Therefore, the net enclosed current is 100[mA] + (-90[mA]) = 10[mA]. This is the current that should be used in the equation for Ampere's Law.

I hope this helps clarify your understanding of the problem. Please let me know if you have any further questions. Good luck with your homework!