Example 37.18. Magnetism in Copper Rod Carrying Current.
A long copper rod of radius \(R\) is carrying a current \(I\) uniformly distributed over the cross-section.
(a) Find the magnitude and direction of the applied field \(\vec H\text{,}\) magnetization \(\vec M\text{,}\) and magnetic field \(\vec B\) inside and outside the wire. Use \(\chi_{M,\text{Cu}}\) for magnetic susceptibility of Copper.
(b) Find numerical values for \(R=1\text{ cm}\) and \(I=100\text{ A}\) at a point \(0.5\text{ cm}\) from the center of the wire.
Data: \(\chi_{M,\text{Cu}} = −0.97\times 10^{-5}\text{.}\)
Answer 1. (a)
(a) Magnitudes at a point a distance \(r\lt R\) from center : \(H_{\textrm{in}} = J\: r/2\text{,}\) \(M = \chi_m\: J\: r/2\text{,}\) \(B_{\textrm{in}} = \mu_0\: (1+\chi_m)\: J\: r/2\text{.}\) Here \(J= I/\pi R^2\text{.}\)
Answer 2. (b)
(b) \(5.0\times 10^{4}\text{ A/m}\text{,}\) \(-4.85\times 10^{4}\text{ A/m}\text{,}\) \(1.88\times 10^{-3}\text{ T}\text{.}\)
Solution 1. (a)
(a) Let \(s\) be the cylindrical radial distance from the center of the wire. Then, applying Ampere’s law to the applied field \(\vec H\) for the cylindrical symmetery, we immediately know that the magnitude \(H\) is
\begin{equation*}
H = \begin{cases}
\dfrac{I}{R^2}\, s, \amp s \lt R\\
\dfrac{I}{2\pi s}, \amp s \gt R
\end{cases}
\end{equation*}
If you don’t know how I got that, you will need to review Ampere’s law in an earlier chapter. The direction of \(\vec H\) will be tangent to circles around the wire, just as it was for \(\vec B\) when we didn’t worry about magnetization of copper itself.
Using susceptibility of Copper for points inside copper and permeability of vacuum at outside points, we get magnitude of \(\vec M\text{.}\)
\begin{equation*}
M = \begin{cases}
\dfrac{\chi_{M,\text{Cu}}\, I}{R^2}\, s, \amp s \lt R\\
0, \amp s \gt R
\end{cases}
\end{equation*}
Since Cu is diamagnetic, its susceptibility \(\chi_{\text{Cu},M} \lt 0\text{.}\) That means, the direction of \(\vec M\) is opposite of the direction of \(\vec H\text{.}\)
Since \(\vec B = \mu_0(\vec H + \vec M)\) we get \(\vec B\) from the results given above. Alternaterly, we can use permeability of Copper for points inside copper and permeability of vacuum at outside points, we get magnitude of \(\vec B\text{.}\)
\begin{equation*}
B = \begin{cases}
\dfrac{\mu_\text{Cu}\, I}{R^2}\, s, \amp s \lt R\\
\dfrac{\mu_0\,I}{2\pi s}, \amp s \gt R
\end{cases}
\end{equation*}
with
\begin{equation*}
\mu_\text{Cu} = 1 + \chi_{M,\text{Cu}}.
\end{equation*}
The direction of \(\vec B\) is same as the direction of \(\vec H\text{.}\)
Solution 2. (b)
(b) The magnitudes of the fields at the indicated point inside are
\begin{align*}
H_\text{in} \amp = \dfrac{I}{R^2}\, s\\
\amp = \dfrac{100\text{ A}}{0.01^2\text{ m}^2}\, 0.05\text{ m} = 5.0\times 10^{4}\text{ A/m}. \\
M \amp = \chi_{M,\text{Cu}} H_\text{in} \\
\amp = −0.97\times 10^{-5} \times 5.0\times 10^{4}\text{ A/m} = -4.85\times 10^{4}\text{ A/m}\\
B \amp = \mu_0 (H + M)\\
\amp = 4\pi\times 10^{-7}\text{ T.m/A}\times (5.0-4.85)\times 10^{4} \text{ A/m}= 1.88\times 10^{-3}\text{ T}.
\end{align*}
