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\begin{document}
    \maketitle
    
    \section{Last Lecture}
    Magnetic force on a wire independent of shape;

    Motion of a charged particle in uniform $\bB$;\\
        Cyclotron radius: $r=\frac{mv}{qB}$ meters\\
        Cyclotron frequency: $\omega = \frac{qB}{m}$ radian Hertz.

    \section{Today}
    \begin{enumerate}
    \setcounter{enumi}{2}
        \item Charged particle motion in magnetic field (last lecture)
        \begin{enumerate}
            \setcounter{enumii}{1}
            \item Lorentz force law
        \end{enumerate}
        \item Torque on current loop -- magnetic dipole moment
        \begin{enumerate}
            \item Torque \& magnetic moment
            \item Potential energy
            \item Gyromagnetic M/L
        \end{enumerate}
        \item Hall Effect
    \end{enumerate}
    
    \section{And the actual notes \ldots}

    Professor McKee was away doing something awesome with a massive telescope 
    probe, which will be in the shadow of the earth to look at the infrared 
    spectrum. They want to find the light from the first stars in the universe.
    Will be launched in June 2015.

    NB that $\frac{mv}{qB} = \frac{p}{qB}$. Neat.

    \setcounter{subsection}{2}

    \subsection{Charged particle in a magnetic field (last lecture)}

    \setcounter{subsubsection}{1}
    \subsubsection{Lorentz force law}

    $\fF_e = q\eE$, $\fF_b = q(\vv \times \bB)$. Total electromagnetic force on
    a charged particle $\fF = q(\eE + \vv \times \bB)$. This holds true even
    when dealing with velocities close to $c$.

    Let's work out an example: what if we put a charged particle in crossed 
    $\eE$ and $\bB$ fields, i.e., $\eE$ perpendicular to $\bB$. If the particle 
    moves at a velocity $\vv = \frac{\eE \times \bB}{B^2}$ $(E << B)$, then 
    $\fF = 0$. That is to say, there's a specific velocity that the particle 
    can move with without any force exerted.

    Remember that $\aA \times (\bB \times \cC) = \bB(\aA \cdot \cC) - \cC
    (\aA \cdot \bB)$; then $$\fF_Bq^{-1} = \vv \times \bB = 
    \frac{(\eE \times \bB) \times \bB}{B^2}$$
    $$ = (\eE \times \hat{B}) \times \hat{B} = -\hat{B} \times (\eE \times 
    \hat{B})$$
    $$ = -[\eE(\hat{B}\cdot\hat{B}) - \hat{B}(\hat{B}\cdot\eE)] = -\eE (\Leftarrow \hat{B} \perp \eE) $$

    \subsection{Torque on Current Loop}
    
    \subsubsection{Magnetic Dipole Moment}

    Recall that $\pp = q\dd, \ttau = \pp \times \eE, U = -\pp \cdot \eE$, we 
    want something similar for a magnetic dipole.

    Consider a planar current loop with top length $b$ and side length $a$ 
    with a current $I$ running right-to-left; an area vector points normal 
    to the plane at some angle $\theta$ to the vertical; $\bB$ is going ``up''.

    First, we know that $\fF = I\lell \times \bB$; forces on the opposite sides
    cancel, since $\lell \parallel I$. But what's the torque? $\ttau = \rr 
    \times \fF$; the roll torque ($b$-length) of the current loop is 0: 
    $\ttau = \rr \times \fF = \frac{\bb}{2} \times \fF = 0$, since 
    $\bb \parallel \fF$.

    How about the pitch? $\tau = \frac{a}{2}IbB\sin\theta$ on each; so 
    $\sum \tau = abIb\sin\theta$. Since $A = ab$, $\tau = IAB\sin\theta$, 
    with vectors, have that $\ttau = I\aA \times \bB$.

    If we have $n$ loops of wire, the total current is $nI$, so the torque 
    becomes $nI\aA \times \bB$.

    Then define the magnetic dipole moment, $\mmu \equiv nI\aA$, so $\ttau = 
    \mmu \times \bB$.

    \subsubsection{Potential energy}

    First, let's record our expectations -- given the diagram as before, 
    the torque will align $\mmu \uparrow \uparrow \bB$.

    Work done by field, $dW = -\tau d\theta$; $dU = -dW = \tau d\theta = 
    \mu B \sin\theta d\theta \Rightarrow U = -\mu B \cos\theta + c$, choose 
    the constant so that $U=0 \longleftrightarrow \theta = \frac{\pi}{2}$,
    so $U = -\mmu \cdot \bB$, implying that the state of maximum energy is 
    when $\mmu \uparrow \downarrow \bB$.

    \subsubsection{Gyromagnetic ratio}

    Gyromagnetic ratio is the ratio of the magnetic dipole moment to the angular
    momentum. They seem totally unrelated, but we'll see that it actually just
    comes down to the ratio of the charge and the mass.

    Take, e.g., a point charge orbiting in uniform $B$. $\mu = IA$, 
    $I = \frac{q}{T}$ ($vT = 2\pi r$). $\Rightarrow I = \frac{qv}{2\pi r}$, 
    so $\mu = IA = \frac{qv}{2\pi r}\pi r^2 = \frac{1}{2}qvr$.

    Now, what's the angular momentum? $L = mrv \rightarrow \frac{\mu}{L} = 
    \frac{1}{2}\frac{qvr}{mrv} = \frac{1}{2}frac{q}{m}$, which is one-half the 
    charge-to-mass ratio.

    This has a direct application to magnetic materials: in the Bohr model of 
    the atom, we have a proton at the center with an electron orbiting it; the
    essence of Bohr's model is that the angular momentum is quantized. 
    $L = \hbar = \frac{h}{2\pi}$; $\mu = \frac{1}{2} \frac{e}{m_e} \hbar$.
    If we put in numbers, $\mu = 9.27\times10^{-24}$ J T$^{-1}$, a 
    typical value for an atomic mgnetic dipole.

    \subsection{Hall Effect}

    Named for the physicist who discovered it in the later part of the 19th 
    century.

    Suppose a wire with a current $I$ flowing to the right in a magnetic field
    flowing away from us; therefore the force $I\lell \times \bB$ is pointing 
    upwards.

    $\fF = q(\eE_{current} + \vv_d \times \bB + \eE_H)$, require that 
    $\fF = q(\eE_{current} \rightarrow q\eE_H = -q\vv_d \times \bB$.

    If wire has diameter $d$, Hall emf = $\xi = E_H d = v_eBd$.
\end{document}