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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} 

{\bf Set 13:} 

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\item{\bf Magnetic field in a hole of a conductor.} \smallskip
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\noindent We choose the axis of the conductor (and the hole) in the 
$\hat{z}$ direction. The current density is $\vec{J}=J\,\hat{z}$ with
$$ J\ =\ \frac{I}{A}\ =\ \frac{I}{\pi\,(R^2-r^2)}\ .$$
The trick is to write the magnetic field as a linear combination
$\vec{B}=\vec{B}_1+\vec{B}_2$ where $\vec{B}_1$ correspond to 
the current density $\vec{J}_1=\vec{J}$ flowing over the {\it entire}
cross section of the conductor, and $\vec{B}_2$ to the current
density $\vec{J}_2=-\vec{J}$ flowing over the cross section of the 
cylindrical hole. Due to the linearity of Amp\'ere's law this give 
the same magnetic field as the original arrangement. With $i=1,\,2$ 
and $\vec{J}_i=J_i\,\hat{z}$ we have
$$ 2\pi\,\rho_i\,B_i\ =\ \oint_C \vec{B}_i\cdot d\vec{l}\ =\ 
   \frac{4\pi}{c}\int_S J_i\,da\ =\ \frac{4\pi^2}{c}\,\rho_i^2\,J_i\,, $$
where the integration contours are circles of radii $\rho_i$ about the 
axis of the conductor ($i=1$) and the cylindrical hole ($i=2$). The
magnetic field vectors are
$$ \vec{B}_i\ =\ \frac{2\pi}{c}\,\rho_i\,J_i\,\hat{\phi}_i\,,\qquad 
   \hat{\phi}_i\ =\ - \sin(\phi_i)\,\hat{x}+\cos(\phi_i)\,\hat{y} $$
where have choosen the center of the hole on the $x$ axis and
defined the angles $\phi_i$ by $\cos(\phi_i)=\vec{\rho}_i\cdot\hat{x}$,
where $\vec{\rho}_i$ are the vectors from the axis of the conductor
($i=1$) and from the axis of the hole ($i=2$) to the (same) point at
which we calculate the magnetic field. The magnetic field becomes
$$ \vec{B}\,=\,\frac{2\pi}{c}\,J\,\{
   \rho_1\,[-\sin(\phi_1)\,\hat{x}+\cos(\phi_1)\,\hat{y}\,] -
   \rho_2\,[-\sin(\phi_2)\,\hat{x}+\cos(\phi_2)\,\hat{y}\,]\} $$
From the geometry we have for the components of the position vector
$$ x = \rho_1\,\cos(\phi_1) = d + \rho_2\,\cos(\phi_2)~~{\rm and}~~
   y = \rho_1\,\sin(\phi_1) = \rho_2\,\sin(\phi_2)\ . $$ 
Therefore,
$$ \vec{B}\,=\, \frac{2\pi}{c}\,J\,d\,\hat{y} \,=\, 
   \frac{2\,I\,d}{c\,(R^2-r^2)}\,\hat{y} $$
holds. In the cylindrical hole there is a constant field in the 
$\hat{y}$ direction.
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