# Magnetic field in coaxial cables

by A_B
Tags: cables, coaxial, field, magnetic
 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 $r_1$, and and outer conducting cylindrical shell of inner radius $r_2$ and outer radius $r_3.$ In both conductors, the current equals $I_0$, 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 $j_1=C_1 r$ and in the outer cable as $j_2=C_2 r$. Calculate the magnetic induction (which I take means magnetic field strength) as function of $R$, the distance from the central axis, and $I_0$ for: a) $R < r_1$ b) $r_1 < R < r_2$ c) $r_2 < R < r_3$ d) $R > r_3$ 2. Relevant equations I think the static Ampère's law is the correct equation $$\oint \bar{B} \cdot d\bar{s} = \mu_0 I$$ 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 $R < r_1$ 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 $$\oint \bar{B} \cdot d\bar{s} = 2\pi B R.$$ For the RHS we need to find the total current through the circle. \begin{align*} 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 \end{align*} If we take this integral from $0$ to $r_1$ , we know that the current through the circle is $I_0$. Thus, we can figure out the constant $C_1.$ $$C_1 = \frac{3I_0}{2\pi r_1^3}$$ We put this in the previous formula for the current to obtain: $$I = \left(\frac{R}{r_1}\right)^3 I_0$$ We plug it all into Ampère's law to obtain $$B=\frac{\mu_0 R^2}{2\pi r_1^3}I_0$$ b) When $r_1 < R < r_2$ the current through the circle is always $I_0$ and the countour integral still equals $2\pi BR$ so the magnetic field strength is $$B = \frac{\mu_0 I_0}{2\pi R}$$ c)The current through the circle is $I_0$ minus the current through the bit of the outer conductor that goes through the circle. We shall first calculate the current through this bit. \begin{align*} 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) \end{align*} Taking this integral to $r_3$ should give $I_0$ so we can figure out the constant $C_2$. $$C_2 = \frac{3I_0}{2\pi \left(r_3^3-r_2^3\right)}$$ Putting it all together gives a total current through the circle of: $$I = I_0 - I_2 = I_0 - \frac{\left(R^3 - r_2^3\right) I_0}{\left(r_3^3 - r_2^3\right)}$$ Applying Amp's law gives $$B = \frac{\mu_0 I_0}{2 \pi R} \left( 1- \frac{R^3 - r_2^3}{r_3^3 - r_2^3} \right)$$ c) The total current is zero, therefore, so is the magnetic field. $$B = 0$$