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

{\bf Set 3:} \bigskip

% \begin{figure}[tb] % \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

{\bf 4.\ Magnetism in matter.}

\begin{enumerate}
\item For this magnetostatics problem with no free currents the relevant
Maxwell equations are
$$ \nabla\cdot\vec{B}=0~~~{\rm and}~~~\nabla\times\vec{H}=0 $$
with the associated boundary conditions (BCs)
$$ \left.\hat{r}\cdot\left(\vec{B}_2-\vec{B}_1\right)\right|_{r=a,b}
   =0\,,~~~\left.\hat{r}\times\left(\vec{H}_2-\vec{H}_1\right)
   \right|_{r=a,b} = 0\,.$$

\item Since $\nabla\times\vec{H}=0$ we have $\vec{H}=-\nabla\Phi$
and thus $0=\nabla\cdot\vec{B}=\nabla\cdot\left(\mu\vec{H}\right)$.

From the BCs we get
$$ 0=\hat{r}\cdot\left(\vec{B}_2-\vec{B}_1\right)=
   \hat{r}\cdot\left(\mu_2\vec{H}_2-\mu_1\vec{H}_1\right)=
   -\mu_2\frac{\partial\Phi_2}{\partial\,r}
   +\mu_1\frac{\partial\Phi_1}{\partial\,r}               $$
and, because of symmetry $\Phi(\vec{x})=\Phi(r,\theta)$,
$$ 0=\hat{r}\times\left(\vec{H}_2-\vec{H}_1\right)
    =\hat{r}\times\left(-\nabla\Phi_2+\nabla\Phi_1\right)
    =\frac{\hat{r}\times\hat{\theta}}{r}\left(
    -\frac{\partial\Phi_2}{\partial\,\theta}
    +\frac{\partial\Phi_1}{\partial\,\theta}\right)\,.     $$

\item The solution is dictated by the behavior of the field 
at infinity:
$$ \lim_{r\to\infty} \vec{B}(\vec{r}) = \vec{B}_0 = B_0\,\hat{z}\
   \Rightarrow\ \lim_{r\to\infty} \Phi(\vec{r}) = - B_0\,z =
   -B_0\,r\,\cos\theta = -B_0\,r\,P_1(\cos\theta)\,.  $$
Thus, we obtain for the $P_l(\cos\theta)$ Legendre polynomials
\begin{eqnarray} \nonumber
  \Phi_1(\vec{r}) &=& A\,r^l\,P_l(\cos\theta)~~{\rm for}\ r<a~~
  ({\rm no\ singular\ contribution})\,,\\ \nonumber
  \Phi_2(\vec{r}) &=& \left(B\,r^l + \frac{C}{r^{l+1}}\right)\,
  P_l(\cos\theta)~~{\rm for}\ a<r<b\,, \\ \nonumber
  \Phi_3(\vec{r}) &=& -\delta^{1l}\,B_0\,r\,\cos\theta+
  \frac{D}{r^{l+1}}\,P_l(\cos\theta)~~{\rm for}\ r>b\,.
\end{eqnarray} 
Using the BCs we obtain:
\begin{eqnarray} \nonumber
  A\,a^l &=& B\,a^l + \frac{C}{a^{l+1}}~~~({\rm continuity\ of}\ 
           \hat{r}\times\vec{H}\ {\rm at}\ r=a)\,, \\ \nonumber
  l\,A\,a^{l-1}  &=& \mu\,\left(l\,B-(l+1)\,\frac{C}{a^{l+2}}\right)
  ~~~({\rm continuity\ of}\ \hat{r}\cdot\vec{B}\ {\rm at}\ r=a)\,, 
  \\ \nonumber
  B\,b^l + \frac{C}{b^{l+1}} &=&-\delta^{1l} B_0\,b + \frac{D}{b^{l+1}
  }~~~({\rm continuity\ of}\ \hat{r}\times\vec{H}\ {\rm at}\ r=b)\,, 
\end{eqnarray} 
\begin{eqnarray} \nonumber
 &~& \mu\,\left(l\,B\,b^{l-1}-(l+1)\,\frac{C}{b^{l+2}}\right)\ = 
 \\ \nonumber &-& \delta^{1l} B_0-(l+1)\,\frac{D}{b^{l+2}}~~~
  ({\rm continuity\ of}\ \hat{r}\cdot\vec{B}\ {\rm at}\ r=b)\,. 
\end{eqnarray} 
For $l\ge 2$ a solution is $A=B=C=D=0$ and it is the only solution as
long as the determinant of the associated linear system of four equations
with four unknowns is not zero. Thus, we keep only $l=1$:
\begin{eqnarray} \nonumber
  A\,a &=& B\,a + \frac{C}{a^2}~~~({\rm continuity\ of}\ 
           \hat{r}\times\vec{H}\ {\rm at}\ r=a)\,, \\ \nonumber
    A  &=& \mu\,\left(B-\frac{2C}{a^3}\right)~~~({\rm continuity\ 
           of}\ \hat{r}\cdot\vec{B}\ {\rm at}\ r=a)\,, \\ \nonumber
  B\,b + \frac{C}{b^2} &=&-B_0\,b + \frac{D}{b^2}~~~({\rm continuity\ 
  of}\ \hat{r}\times\vec{H}\ {\rm at}\ r=b)\,, \\ \nonumber
  \mu\,\left(B-\frac{2C}{b^3}\right) &=& -B_0-\frac{2D}{b^3}~~~
  ({\rm continuity\ of}\ \hat{r}\cdot\vec{B}\ {\rm at}\ r=b)\,. 
\end{eqnarray} 
From the BCs at $r=a$ we obtain
\begin{eqnarray} \nonumber
  B = \left(\frac{1+2\mu}{3\mu}\right)\,A\,,~~~
  C = \left(\frac{\mu-1}{3\mu}\right)\,a^3\,A\,,
\end{eqnarray} 
and from the BCs at $r=b$ we get
\begin{eqnarray} \nonumber
  B = -\left(\frac{1+2\mu}{3\mu}\right)\,B_0 
   + \frac{2\,(\mu-1)}{3\mu\,b^3}\,D \,,~~~
  C = -\left(\frac{\mu-1}{3\mu}\right)\,b^3\,B_0 
   + \left(\frac{2+\mu}{3\mu}\right)\,D \,.
\end{eqnarray} 
Thus, the four unknown constant can all be determined in terms of $B_0$. 
Eliminating $B$ and $C$ in favor of $A$ in the last two equations, 
these become
\begin{eqnarray} \nonumber
  \left(\frac{1+2\mu}{3\mu}\right)\,A &=&
   - \left(\frac{1+2\mu}{3\mu}\right)\,B_0 
   + \frac{2\,(\mu-1)}{3\mu\,b^3}\,D\,,\\ \nonumber
  \left(\frac{\mu-1}{3\mu}\right)\,a^3\,A &=&
   - \left(\frac{\mu-1}{3\mu}\right)\,b^3\,B_0 
   + \left(\frac{2+\mu}{3\mu}\right)\,D \,,
\end{eqnarray} 
with the solutions
\begin{eqnarray} \nonumber
  A = \frac{-9\,b^3\,\mu\,B_0}{b^3\,(2\mu+1)\,(\mu+2)-2a^3\,(\mu-1)^2}
  \,,~~~ D = \frac{(2\mu+1)\,(\mu-1)\,(b^3-a^3)\,b^3\,B_0}
                  {b^3\,(2\mu+1)\,(\mu+2)-2a^3\,(\mu-1)^2}\,.
\end{eqnarray} 
The constants $B$ and $C$ for the region $a<r<b$ are best expressed
in terms of $A$. That is,
\begin{eqnarray} \nonumber
  B = \left(\frac{1+2\mu}{3\mu}\right)\,A\,,~~~ 
  C = \left(\frac{\mu-1}{3\mu}\right)\,a^3\,A\,.
\end{eqnarray} 
Note that for $B_0=0$ all coefficient become zero. \medskip

\item In the limit $\mu\to\infty$ we obtain $A\sim B_0/\mu$ and, hence,
$A=B=C=0$ (no field in the inner sphere) and
$$ D\ \to\ \frac{2\mu^2\,(b^3-a^3)\,b^3\,B_0}{2\mu^2\,(b^3-a^3)}
   \ =\ b^3\,B_0\,.$$
Thus,
\begin{eqnarray} \nonumber
  \Phi_1(\vec{r})\ =\ 0~~~{\rm and}~~~\Phi_3(\vec{r}) &=&
  \left(-B_0\,r+\frac{b^3}{r^2}\,B_0\right)\,\cos\theta\\ \nonumber 
  &=& -B_0\,z+\frac{\vec{m}\cdot\vec{r}}{r^3}~~{\rm with}~~
      \vec{m}=b^3\,B_0\,\hat{z}\,.
\end{eqnarray} 
We obtain
\begin{eqnarray} \nonumber
  B_3(\vec{r})&=&-\nabla\Phi_3(\vec{r})\ =\ B_0\,\hat{z} -
  (\vec{m}\cdot\vec{r}\,)\,\nabla\,\frac{1}{r^3} -
  \frac{1}{r^3}\,\nabla\,(\vec{m}\cdot\vec{r}\,) \\ \nonumber
  &=& \vec{B}_0+\frac{3\,(\vec{m}\cdot\vec{r}\,)\,\hat{r}}{r^4} 
   - \frac{\vec{m}}{r^3}\ =\ \vec{B}_0 +
   \frac{3\,(\vec{m}\cdot\hat{r}\,)\,\hat{r}-\vec{m}}{r^3}\,,
\end{eqnarray} 
i.e., we find a magnetic dipole correction to the constant magnetic 
field.
\end{enumerate}


{\bf 5.\ Faraday's law of induction in a constant magnetic field.} 
\smallskip % DONE: added to solutions appendix.

We use the initial condition $\hat{n}=\hat{y}$ at time $t=0$, where 
$\hat{n}$ is the normal to the surface. \smallskip

\begin{enumerate}

\item For the left-hand side we have to calculate
$$ -\frac{1}{c}\,\frac{\Phi_m}{d\,t}~~{\rm with}~~
   \Phi_m\ =\ \int_S \vec{B}\cdot\hat{n}\,da~~{\rm and}~~\hat{n}
   = \hat{\phi}=-\sin(\omega\,t)\,\hat{x}+\cos(\omega\,t)\,\hat{y}$$
for the normal to the surface, where we use cylindrical coordinates 
and $\phi=\omega t$. With $\vec{B}=B_0\,\hat{y}$ we obtain for 
the left-hand side the result
\begin{eqnarray} \nonumber
  \vec{B}\cdot\hat{n} = B_0\,\cos(\omega\,t)~&{\rm and}&~
  \Phi_m = B_0\,L^2\,\cos(\omega\,t)\,,\\ \nonumber 
  -\frac{1}{c}\,\frac{d~}{dt}\,\Phi_m &=& 
  c^{-1}B_0\,L^2\,\omega\,\sin(\omega\,t)\,.
\end{eqnarray}
On the right-hand side the electric field $\vec{E}$ does not contribute,
because we have a constant magnetic field and, hence, 
\begin{eqnarray} \nonumber
  \nabla\times\vec{E} = -\frac{1}{c}\,\frac{\partial~}{\partial t}\,
  \vec{B} = 0\ \Rightarrow\ \oint_C \vec{E}\cdot d\vec{l} = 0\,,
\end{eqnarray}
so that the equation becomes
\begin{eqnarray} \nonumber
  \frac{1}{c}\oint_C\left(\vec{v}\times\vec{B}\right)\cdot d\vec{l}\,,
\end{eqnarray} 
where the velocity is that of the line element $d\vec{l}$. With 
$\rho=\sqrt{x^2+y^2}$ 
\begin{eqnarray} \nonumber
  \vec{v} &=& \rho\,\omega\,\hat{\phi} =\rho\,\omega\,\left[ -
  \sin(\omega\,t)\,\hat{x}+\cos(\omega\,t)\,\hat{y} \right]\\ \nonumber
  \vec{v}\times\vec{B}&=&-\rho\,\omega\,B_0\,\sin(\omega\,t)\,\hat{x}
  \times\hat{y}\ =\ -\rho\,\omega\,B_0\,\sin(\omega\,t)\,\hat{z}\,.
\end{eqnarray} 
Therefore, only the line elements in $\hat{z}$ direction contribute,
which are at $\rho = L/2$. The integral becomes (note the right-handed
orientation of the loop)
\begin{eqnarray} \nonumber
  \frac{1}{c}\oint_C \left(\vec{v}\times\vec{B}\right)\cdot d\vec{l}
  &=&-\frac{B_0\,L\,\omega}{2\,c}\,\sin(\omega\,t)\,\left[ \int_L^0 dz 
   -\int_0^L dz\,\right]\\ \nonumber 
  &=& c^{-1}B_0\,L^2\,\omega\,\sin\omega\,t)\,.
\end{eqnarray} 
\smallskip

\item We have
$$ \nabla\cdot\vec{v} =\left(\hat{\rho}\,\frac{\partial~}{\partial\rho}
  +\frac{\hat{\phi}}{\rho}\,\frac{\partial~}{\partial\phi}+
  \hat{z}\,\frac{\partial~}{\partial z}\right)\cdot \omega\,\rho\,
  \hat{\phi} = 0~~\Rightarrow~~\frac{1}{c} \int_S 
  \left(\nabla\cdot\vec{v}\right)\,\vec{B}\cdot d\vec{a} = 0\,. $$
Next, 
\begin{eqnarray} \nonumber
  \left(\vec{B}\cdot\nabla\right)\,\vec{v} &=& B_0\,\frac{\partial
  \vec{v}}{\partial y}\ =\ B_0\,\hat{y}\cdot\left(\hat{\rho}\,
  \frac{\partial~}{\partial\rho}+\frac{\hat{\phi}}{\rho}\,
  \frac{\partial~}{\partial\phi}+\hat{z}\, 
  \frac{\partial~}{\partial z}\right)\,\vec{v} \\ \nonumber &=&
  B_0\,\left(\sin(\omega t)\,\frac{\partial~}{\partial\rho} +
  \frac{\cos(\omega t)}{\rho}\,\frac{\partial~}{\partial\phi}\right)\,
  \omega\,\rho\,\hat{\phi}\ \\ \nonumber &=&\ B_0\,\omega\,
  \left(\sin(\omega t)\,\hat{\phi} - \cos(\omega t)\,\hat{\rho}\right) 
\end{eqnarray} 
and, therefore (using $\hat{\phi}\cdot\hat{n}=1$ and $\hat{\rho}\cdot
\hat{n}=0$),
$$ \frac{1}{c} \int_S \left(\vec{B}\cdot\nabla\right)\,\vec{v}\cdot
   d\vec{a} \ =\ c^{-1}\omega\,B_0\,L^2\,\sin(\omega t)\,.        $$
Further
$$ \frac{1}{c} \int_S \left(\vec{v}\cdot\nabla\right)\,\vec{B}\cdot
   d\vec{a}\ =\ 0\,, $$
because $\vec{B}$ is constant. Finally,
\begin{eqnarray} \nonumber
  \vec{v}\times\vec{B}&=&-\rho\,\omega\,B_0\,\sin(\omega t)\,\hat{z}\,,
  \\ \nonumber
  \nabla\times \rho\,\omega\,B_0\,\sin(\omega t)\,\hat{z} &=&
   -\omega\,B_0\,\sin(\omega t)\,\hat{\rho}\times \hat{z}\ =\ +
   \omega\,B_0\,\sin(\omega t)\,\hat{\phi} - 
   \omega\,B_0\,\sin(\omega t)\,\hat{\rho}
\end{eqnarray} 
and $\hat{\phi}\cdot\hat{n}=1$ implies
$$ \frac{1}{c} \int_S \nabla\times \left(\vec{v}\times\vec{B}\right)
   \cdot d\vec{a} \ = \ c^{-1}B_0\,L^2\,\omega\,\sin(\omega\,t) $$
as before. The results are consistent with the vector relation
$$ \nabla\times \left(\vec{v}\times\vec{B}\right)\ =\
   \left(\nabla\cdot\vec{B}\right)\,\vec{v} +
   \left(\vec{B}\cdot\nabla\right)\,\vec{v} -
   \left(\nabla\cdot\vec{v}\right)\,\vec{B} -
   \left(\vec{v}\cdot\nabla\right)\,\vec{B} \,.  $$ 
\bigskip 

\item We have now $\hat{n}=\hat{x}$ and, therefore,
$$ \Phi_m\ =\ \int_S \vec{B}\cdot\hat{n}\,da\ =\ B_0\,L^2\,\sin(\omega\,t)\,.$$
Differentiation with respect to the time gives
$$ -\frac{1}{c}\,\frac{d~}{dt}\,\Phi_m =-c^{-1}B_0\,L^2\,\omega\,
    \cos(\omega\,t)\,.  $$
This is the previous result phase-shifted due to different initial
conditions.
\bigskip %  \vfil\eject
\end{enumerate}


{\bf 6.\ LCR circuit.} \smallskip

\begin{enumerate}
  \item As function of the angular frequency the maximum of the current 
  is given by
  \begin{eqnarray} \nonumber
    I_{\max}(\omega) = \frac{\epsilon_{\max}}{Z} = \frac{\epsilon_{\max}}
    {\sqrt{\left[1/(\omega C)-\omega L\right]^2+R^2}}\,,
  \end{eqnarray} 
  which implies for the resonance frequency
  \begin{eqnarray} \nonumber
    f_0 = \frac{\omega_0}{2\pi} = \frac{1}{2\pi\sqrt{LC}} =
    \frac{1}{2\pi\sqrt{2\,[H]\,2\,10^{-6}\,[F]}} =
    \frac{1}{4\pi\,10^{-3}\,[s]} = 79.6\,[Hz]\,.
  \end{eqnarray} 

  \item Let $\epsilon_{\max}=100\,[V]$ and $\omega_{60} = 120\pi\,[Hz]$. 
  Then,
  \begin{eqnarray} \nonumber
  X_C\ =\ \frac{1}{\omega_{60}C} &=& 1326.3\,[V/A]\,,~~~~
  X_L\ =\ \omega_{60}L\ =\ 754.0\,[V/A]\,,\\ \nonumber
  I_{\max}(\omega_{60}) &=&
  \frac{100\,[V]}{\sqrt{\left(X_C-X_L\right)^2+R^2}}=0.175\,[A]\,,
  \end{eqnarray} 
  whereas at resonance frequency we have $I_{\max}(\omega_0)=
  \epsilon_{\max}/R=5\,[A]$. This is with 28.6 almost thirty times larger 
  than the maximum for $\omega_{60}$.

  For the charge maximum we have 
  \begin{eqnarray} \nonumber
    Q_{\max}(\omega) = \frac{\epsilon_{\max}}{\omega\,Z} 
    = \frac{I_{\max}(\omega)}{\omega}\,.
  \end{eqnarray} 
  resulting in $Q_{\max}(\omega_{60})=0.00046\,[C]$ and $Q_{\max}
  (\omega_0)=0.01\,[C]$. The ratio is down to 21.6.

  \item The phase shift at $60\,[Hz]$ is rather small:
  \begin{eqnarray} \nonumber
  \tan\delta\ =\ \frac{R}{X_L-X_C}\ =\ -0.0349\ \Rightarrow\
  \delta\ =\ -2^o\,.
  \end{eqnarray} 
\end{enumerate}
\bigskip

{\bf 7.\ Complex numbers and integration.} \smallskip

\begin{enumerate}
  \item $\overline{z}=x-iy$. \medskip

  \item $|z|=+\sqrt{x^2+y^2}$.  \bigskip
  \item $$~~~~~$$ \vskip -70pt
  $$\frac{1}{z}=\frac{\overline{z}}{z\,\overline{z}}=\frac{x-iy}
  {x^2+y^2}\,.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ 
  \item $z=|z|\,e^{i\phi}=r\,e^{i\phi},~~\tan(\phi)=y/x$.
  \item In cylindrical coordinates
   $$ I_n = \oint_C dz\,z^n = \int_0^{2\pi} R\,i\,e^{i\,\phi}\,d\phi\,
      R^n\,e^{i\,n\,\phi} = i\,R^{n+1} \int_0^{2\pi} d\phi\,e^{i\,(n+1)
      \,\phi} $$
   For $n\ne -1$:
  $$ I_n = \frac{i\,R^{n+1}}{i\,(n+1)}\, \left. 
     e^{i\,(n+1)\,\phi}\right|_0^{2\pi} = \frac{R^{n+1}}{(n+1)}\,
     \left(e^{i\,(n+1)\,2\pi}-1\right)=0\,.$$ 
  For $n=-1$:
  $$ I_{-1} = \oint_C \frac{dz}{z} = 
     i\,\int_0^{2\pi} d\phi = 2\pi\,i\,. $$
\end{enumerate}
\bigskip

\end{document}
