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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \medskip

{\bf Set 1:} \medskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{description}
\item{\bf 1. Solenoid.}
\end{description}

\begin{enumerate}
\item For a single loop in the $x$--$y$ plane, centered
at the origin, we have derived
% (\ref{A_circular_loop})
$$ \vec{A}=A^{\phi}\,\hat{\phi}~~{\rm with}~~A^{\phi}=
   \frac{I\,R}{c}\int_0^{2\pi} \frac{\cos\phi'\,d\phi'}
   {\sqrt{R^2+r^2-2R\,r\,\sin\theta\,\cos\phi'}}\ . $$
In the cylindrical coordinates $A^{\phi}$ reads
$$ A^{\phi}= \frac{I\,R}{c}\int_0^{2\pi} \frac{\cos\phi'\,d\phi'}
   {\sqrt{R^2+z^2+\rho^2-2R\,\rho\,\cos\phi'}}\ . $$
For $\rho\ll R$ the expansion to order $\rho/R$ is
\begin{eqnarray} \nonumber
  A^{\phi} &=& \frac{I\,R}{c} \left[ 
  \int_0^{2\pi} \frac{\cos\phi'\,d\phi'}{\sqrt{R^2+z^2}} + 
  \int_0^{2\pi} \frac{R\,\rho\,\cos^2\phi'\,d\phi'}{(R^2+z^2)^{3/2}}
  \right] + O\left(\frac{\rho^2}{R^2}\right)\\ \nonumber
  &=& \frac{\pi\,I\,R^2\,\rho}{c\,(R^2+z^2)^{3/2}}~~{\rm as}~~~
  \int_0^{2\pi} \cos^n\phi'\,d\phi' \,=\, \begin{cases} 
  0~~{\rm for}~~n=1\,;\\ \pi~~{\rm for}~~n=2\,. \end{cases}
\end{eqnarray}
If the loop has its center not at the origin but at $z'$ on the
$z$-axis, this equation becomes
$$  A^{\phi}\ =\ \frac{\pi\,I\,R^2\,\rho}{c\,(R^2+(z-z')^2)^{3/2}}\,.$$
In this approximation the potential of a solenoid of length $L$ with 
$n+1$ loops is given by
\begin{eqnarray} \nonumber
  A^{\phi}&=&\sum_{i=0}^n \frac{\pi\,I\,R^2\,\rho}{c\,(R^2+(z-z'_i
  )^2)^{3/2}}\\ \nonumber &=&\frac{n}{L} \sum_{i=0}^n 
  \frac{L}{n}\,\frac{\pi\,I\,R^2\,\rho}{c\,(R^2+(z-z'_i)^2)^{3/2}}
  ~~{\rm with}~~z'_i = -\frac{L}{2}+\frac{i\,L}{n} 
\end{eqnarray}
For large $n$ we obtain
\begin{eqnarray} \nonumber
   A^{\phi} &\approx& N\int_{-L/2}^{+L/2} \frac{dz'\,\pi\,I\,R^2\,\rho}
   {c\,(R^2+(z-z')^2)^{3/2}}\\ 
   \nonumber &=& \frac{\pi\,N\,I\,R^2\,\rho}{c} 
   \int_{-L/2-z}^{+L/2-z} \frac{dz''}{c\,(R^2+z''^2)^{3/2}}\,,
\end{eqnarray}
where $N=n/L$ is the number of loops per unit length and we substituted
$z'=z''+z$ in the last step. 

The last integral is elementary
$$ \int^z \frac{dz''}{(R^2+z''^2)^{3/2}}\ =\ 
   \frac{z}{R^2\,\sqrt{R^2+z^2}} + {\rm const}\,, $$
implying
$$ A^{\phi}\ =\ \frac{\pi\,N\,I\,\rho}{c} \left[
   \frac{L/2-z}{\sqrt{R^2+(L/2-z)^2}} -
   \frac{-L/2-z}{\sqrt{R^2+(L/2+z)^2}} \right]\ . $$
From this expression we calculate the magnetic field. First, the 
$z$ component is 
\begin{eqnarray} \nonumber
  B^z &=& \frac{1}{\rho}\frac{\partial~}{\partial\rho}\,
  \left(\,\rho\,A^{\phi}\,\right)\ =\ \frac{2\pi\,N\,I}{c} 
  \left[ \frac{L/2-z}{\sqrt{R^2+(L/2-z)^2}} +
   \frac{L/2+z}{\sqrt{R^2+(L/2+z)^2}} \right]\,.
\end{eqnarray}
To expand for $R,\,z \ll L$ we factor $L/2$ out of the square root
and drop the terms of $1/L^2$ in the square root
\begin{eqnarray} \nonumber
  B^z &=& \frac{4\pi\,N\,I}{c\,L}\, \left[ \frac{L/2-z}
  {\sqrt{1-4z/L+4z^2/L^2+4R^2/L^2}} + \frac{L/2+z}
  {\sqrt{1+4z/L+4z^2/L^2+4R^2/L^2}}\right] \\ \nonumber &=&
  \frac{4\pi\,N\,I}{c\,L}\,\left[\frac{L/2-z}{\sqrt{1-4z/L}} 
  + \frac{L/2+z}{\sqrt{1+4z/L}}\right]\,.
\end{eqnarray}
Finally, we use the following approximations to order $1/L$
$$ \sqrt{1\pm 4z/L}  = 1 \pm \frac{2z}{L}\,,\qquad
   \frac{1}{1\pm 2z/L} = 1 \mp \frac{2z}{L} $$
with the result
$$ B^z = \frac{4\pi\,N\,I}{c}\,\left[ 1 + 
         O\left(\frac{1}{L^2}\right)\,\right] \approx
         \frac{4\pi\,N\,I}{c}\ . $$
Similarly we evaluate the $B^{\rho}$ component 
$$  B^{\rho}\ =\ - \frac{\partial A^{\phi}}{\partial z} 
    \ =\ \frac{\pi\,N\,I\,R^2\,\rho}{c} \left[
    \frac{-1}{[R^2+(L/2+z)^2]^{3/2}} +
    \frac{1}{[R^2+(L/2-z)^2]^{3/2}} \right]\ . $$
We factor $L/2$ out 
\begin{eqnarray} \nonumber
  B^{\rho} \,=\, \frac{8\pi\,N\,I\,R^2\,\rho}{c} &~& \left[\,
    \frac{-1}{L^3\,[1+4z/L+4z^2/L^2+4R^2/L^2]^{3/2}} \right.
  \\ \nonumber &~& + \left. 
    \frac{1}{L^3\,[1+4z/L-4z^2/L^2+4R^2/L^2]^{3/2}}\,\right]\ . 
\end{eqnarray}
and expand for $R,\,z \ll L$ using
$$ (1\pm 4z/L)^{3/2} \,=\, 1 \pm 6\,\frac{z}{L} $$ 
and find
\begin{eqnarray} \nonumber
  B^{\rho} &=& \frac{8\pi\,N\,I\,R^2\,\rho}{c} \left[\,
  \frac{-1}{L^3\,[1+4z/L]^{3/2}} + \frac{1}{L^3\,[1-4z/L]^{3/2}}
  \right]\,\left[ 1+O\left(\frac{1}{L^2}\right) \right]\\ \nonumber 
  &\approx& \frac{8\pi\,N\,I\,R^2\,\rho}{c} \left[\,\frac{12\,z}{L^4}\,
  \right] \,=\, \frac{96\,\pi\,N\,I\,R^2\,z\,\rho}{c\,L^4}\ .
\end{eqnarray}
\item For $z=L/2$ we have
\begin{eqnarray} \nonumber
  B^{\rho} &=& \frac{\pi\,N\,I\,R^2\,\rho}{c} \left[
  \frac{-1}{(R^2+L^2)^{3/2}} + \frac{1}{R^3} \right]\\ \nonumber
  &=& \frac{\pi\,N\,I\,R^2\,\rho}{c}\,\left(\frac{1}{R^3}\right)\,
  \left[ O\left(\frac{1}{L^2}\right) + 1 \right] \approx 
  \frac{\pi\,N\,I\,\rho}{c\,R}\ .
\end{eqnarray}
and
$$ B^z = \frac{2\pi\,N\,I\,R^2}{c} \left[ -
   \frac{-L}{R^2\,\sqrt{R^2+L^2}} \right] = 
   \frac{2\pi\,N\,I}{c}\,\left[1+O\left(\frac{1}{L^2}\right)\right] 
    \approx \frac{2\pi\,N\,I}{c}\ . $$
\end{enumerate}

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