- #1
satchmo05
- 108
- 0
1. 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.
2. Homework Equations
Cylindrically symmetric current distribution - able to use Ampere's Law here!
∫B•dl = μo*Ienc
3. The Attempt at a Solution
Implementing the cylindrically symmetric current distribution, the left side of the equation ends up being:
2∏ρ*BФ(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!
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.
2. Homework Equations
Cylindrically symmetric current distribution - able to use Ampere's Law here!
∫B•dl = μo*Ienc
3. The Attempt at a Solution
Implementing the cylindrically symmetric current distribution, the left side of the equation ends up being:
2∏ρ*BФ(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!