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\it
\noindent
\parbox{7.5in}{Physics for Scientists and Engineers II 
\hfill Phy184, MSU Spring 1997 \\ Lecture Notes 6 \hfill -- mrt --}
\vspace{0.5cm}

\rm
\centerline{MAGNETIC MATERIALS}
\vspace{0.1cm}

\begin{itemize}
  \item Microscopic origin:
          Atoms have small magnetic moments (elementary magnetic dipoles).
          In ferromagnetic materials (iron, nickel $\ldots$) these magnetic
                moments interact so strongly that the dipoles spontaneously
                align.
          Thus the material forms one large magnetic dipole $\rightarrow$
                magnet.
  \item Inserting materials into a magnetic field can polarize the material
        and create a magnetic dipole moment. The total magnetic field can then be much stronger as it is the sum of the external field and the magnetisation of the ferromagnetic material: $B = B_{ext} + B_M$.
\end{itemize}
                                             


\rm
\vspace{0.3cm}
\centerline{AMPERE-MAXWELL'S LAW AND DISPLACEMENT CURRENT}
\vspace{0.1cm}

\begin{itemize}
  \item Just as a changing magnetic flux creates an electric field (Faraday's law, see page 10 of the lecture notes) a changing electric flux can create a  magnetic field:  $\oint \vec B d \vec s = \mu_0 \epsilon_0 d\Phi_E/dt$ (Maxwell's law).
\item Thus, a magnetic field can be created by a current or by a changing electric field, which combines Ampere's and Maxwell's law: 
$\oint \vec B d \vec s = \mu_0 \epsilon_0 d\Phi_E/dt + \mu_0 i_{enc}$.
\item A displacement current can be defined as $i_d = \epsilon_0 d\Phi_E/dt$,
and then the Ampere-Maxwell law can be written as 
$\oint \vec B d \vec s = \mu_0 i$, with $i = i_{enc} + i_d$.
\end{itemize}

\rm
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\centerline{GAUSS' LAW FOR MAGNETIC FIELDS}
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\begin{itemize}
\item In analogy to Gauss' law for electric fields, it also can be written
for magnetic fields: $\Phi_B =  \oint \vec B d \vec A = 0$.
\item This means that the net magnetic flux through an enclosed surface is always zero, or: A magnetic monopole does not exist.
\end{itemize}


\rm
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\centerline{MAXWELL'S EQUATIONS}
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\begin{itemize}
\item The basic laws combining electricity and magnetism are Maxwell's equations:
\begin{center}
\begin{tabular}{|l|l|}\hline
$\oint \vec E d \vec A = q/\epsilon_0$ & $\oint \vec B d \vec A = 0$ \\ \hline
$\oint \vec E d \vec s = - d\Phi_B/dt$ & $\oint \vec B d \vec s = \mu_0 \epsilon_0 d\Phi_E/dt + \mu_0 i_{enc}$ \\ \hline
\end{tabular}
\end{center}
\end{itemize}


\rm
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\centerline{LC OSCILLATOR}
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\begin{itemize}
  \item The total energy stored in an LC oscillator is the sum of the energy stored in the capacitor \\ $U_E = q^2/(2C)$ and the energy stored in the inductor $U_B = i^2L/2$.
  \item The total energy is conserved and thus $dU/dt = 0$.
  \item The solution to the resulting differential equation is: $q = Q cos(\omega t +\phi)$ and $i = I sin(\omega t +\phi)$, where $I = \omega Q$ and $\omega = 1/\sqrt{LC}$.
  \item The total energy is $U = Q^2/(2C)$, $U_E = Q^2/(2C) \cos^2(\omega t + \phi)$ and $U_B = Q^2/(2C) \sin^2(\omega t + \phi)$.
        circuit.
\end{itemize}

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\it
\noindent
\parbox{7.5in}{\hfill Physics for Scientists and Engineers II,
phy184, Spring 1997, page 12}
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\rm
\centerline{RLC OSCILLATOR}
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\begin{itemize}
  \item Now energy is dissipated in the resistor and thus $ dU/dt = -i^2R$.\\
  The solution is now: $q = Q e^{-Rt/2L} \cos(\omega't + \phi)$ with
	$\omega' = \sqrt{\omega^2 - (R/2L)^2}$.\\
   The total energy decreases as $U = Q^2/(2C) e^{-Rt/L}$.
\end{itemize}

\rm
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\centerline{R, L, OR C IN AN AC CIRCUIT}
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\begin{itemize}
\item
If we connect a resistor, a capacitor, or an inductor to an external ac (alternating current) emf, ${\cal E} = {\cal E}_m \sin (\omega_d t)$, $\omega_d$ is called the driving frequency. The voltage drop across the element (x = R, C or L) is then $v_x = V_x \sin (\omega_d t)$.

\item The current can then be calculated with $ i_x = I_x \sin (\omega_d t - \phi)$, where $\phi$ corresponds to the phase difference between the voltage and the current.
\item The equivalence of a resistance for a capacitor and an inductor is the
reactance. The capacitive reactance is $X_C = 1/(\omega_d C)$ and the inductive reactance is $X_L = \omega_d L$.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}\hline
Element & Symbol & Resistance & Phase of & Phase angle & Amplitude \\
	&	 & or Reactance & Current & $\phi$ & Relation \\ \hline
Resistor & R & R & in phase & $0^\circ $ & $V_R = I_R R$ \\
Capacitor & C & $X_C = 1/(\omega_d C)$ & leads $v_C$ (ICE) & $-90^\circ $ & $V_C = I_C X_C$ \\
Inductor & L & $X_L = \omega_d L$ & lags $v_L$ (ELI) & $90^\circ $ & $V_L = I_L X_L$ \\ \hline
\end{tabular}
\end{center}
\end{itemize}

\rm
\vspace{0.3cm}
\centerline{RLC CIRCUIT}
\vspace{0.1cm}

\begin{itemize}
\item In an RLC circuit the instantaneous voltages have to add up to the emf:
${\cal E} = v_R + v_L + v_C$.
\item The amplitudes can be calculated from the vector sum of the phasors.
Thus the maximum current is given by $I = {\cal E}_m/Z$ where $Z = \sqrt{R^2+(X_L - X_C)^2}$ is the impedance of the circuit.
\item The phase constant is defined as $\tan \phi = (X_L - X_C)/R$.
\item The current has a maximum when the impedance has a minimum, i.e. $Z = R$. This occurs at the resonance frequency $\omega_d = \omega = 1/\sqrt{LC}$.
\item The average power dissipated in an ac circuit is $P_{av} = I^2_{rms}R$,
where the rms current is $I_{rms} = I/\sqrt{2}$.
The definition of an rms emf as ${\cal E}_{rms} = {\cal E}_m/\sqrt{2}$ yields also: $P_{av} = {\cal E}_{rms} I_{rms} R/Z = {\cal E}_{rms} I_{rms} \cos \phi$.
\end{itemize}

\rm
\vspace{0.3cm}
\centerline{TRANSFORMERS}
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\begin{itemize}
  \item Transformers consist of two coils (primary and secondary) wound on the same
        iron core with different number of turns.  The same magnetic flux is then in both coils:\\
        $
          V_p = -N_p \Delta \phi_B/ \Delta t $ and
          $ V_s = -N_s\Delta \phi_B/ \Delta t
          \rightarrow V_s/V_p = N_s/N_p
        $. \\
The currents are then related via: $I_s/I_p = N_p/N_s$.
  \item The equivalent resistance of the coil in the primary circuit is given by $R_{eq} = (N_p/N_s)^2 R$, where $R$ is the resistance in the circuit of the second coil.
\end{itemize}



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