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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.
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
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.
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.
<br /> \begin{align*}<br /> I &= \iint\limits_{circle} j_1 dA<br /> \\ &= \int_0^{2\pi} d\theta \int_0^R C_1 r^2 dr<br /> \\ &= \frac{2}{3} \pi C_1 R^3<br /> \end{align*}<br />
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.
<br /> \begin{align*}<br /> I_2 &= \iint\limits_{r_2 \rightarrow R} j_2 dA<br /> \\ &= \int_0^{2\pi} d\theta \int_{r_2}^{R} C_2 r^2 dr<br /> \\ &= \frac{2}{3} \pi C_2 \left(R^2 - r_2^3\right)<br /> \end{align*}<br />
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
Homework Statement
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
Homework 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.
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.
<br /> \begin{align*}<br /> I &= \iint\limits_{circle} j_1 dA<br /> \\ &= \int_0^{2\pi} d\theta \int_0^R C_1 r^2 dr<br /> \\ &= \frac{2}{3} \pi C_1 R^3<br /> \end{align*}<br />
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.
<br /> \begin{align*}<br /> I_2 &= \iint\limits_{r_2 \rightarrow R} j_2 dA<br /> \\ &= \int_0^{2\pi} d\theta \int_{r_2}^{R} C_2 r^2 dr<br /> \\ &= \frac{2}{3} \pi C_2 \left(R^2 - r_2^3\right)<br /> \end{align*}<br />
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