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Magnetic field in coaxial cables 
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#1
May511, 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. [tex] \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*} [/tex] 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. [tex] \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*} [/tex] 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_3^3r_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] 


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