Chapter 12, Problem 81AP

A long, straight, cylindrical conductor contains a cylindrical cavity whose axis is displaced by n from the axis of the conductor, as shown in the accompanying figure. The current density in the conductor is given by

$\stackrel{\to }{J}=J_0\hat{k}$, where $J_0$ is a constant and $\hat{k}$ is along the

axis of the conductor. Calculate the magnetic field at an arbitrary point P in the cavity by superimposing the field of a solid cylindrical conductor with radius R₁and current density $\stackrel{\to }{J}$onto the field of a solid cylindrical conductor with radius R₂and current density $-\stackrel{\to }{J}$ . Then use the fact that the appropriate azimuthal unit vectors can be expressed as ${\hat{\theta }}_1=\hat{k}\times {\hat{r}}_1$and ${\hat{\theta }}_2=\hat{k}\times {\hat{r}}_2$ to show that everywhere inside the cavity the magnetic field is given by the constant $B=\frac{1}{2}{\mu }_0{J}_{0}k\times a$ , where $a=r_1-r_2$ and $r_1=r_1{\hat{r}}_1$ is the position of P relative to the center of the conductor and $r_2=r_2{\hat{r}}_2$ is the position of P relative to the center of the cavity.