Subfield: Electromagnetism

Two small balls of mass $m$ each with charges $+q$ and $-q$ respectively, connected by a rigid massless rod of length $d$, form a dipole. The dipole is parallel to plane $X Y$ and is placed in a uniform magnetic field $\vec{B}$ perpendicular to $X Y$.

<img_4351>

Initially, the dipole is aligned with the direction $X$ and has initial angular velocity $\omega_{0}$ in plane $X Y$, as shown. Its center of mass is initially located at origin and given initial velocity $\vec{v}_{0}$ parallel to $X Y$, as well.

Consider three distinct scenarios (a, b, c-d):
Context question:
(a) Find $\omega_{0}$ and the direction of $\vec{v}_{0}$, so that the center of mass will move with the constant velocity $\vec{v}=\vec{v}_{0}$ ?
Context answer:
$\vec{v}_{0} \| Y$ , $\omega_{0}=0$


Context question:
(b) Given $\omega_{0}$, find such $\vec{v}_{0}$ (direction and magnitude), so that the center of mass will travel in a circle. Find the circle radius $R_{c}$ and the coordinates $x_{c}, y_{c}$ of its center. You don't need to prove the uniqueness of the solution.
Context answer:
$\vec{v}_{0}=-\frac{q d B}{2 m} \hat{\jmath}$ and $R_{c}=\frac{q B d}{2 m|\omega_{0}|}$ and $(x_{c}, y_{c})=( \pm R_{c}, 0)$


Context question:
(c) Given $\vec{v}_{0}=0$, find the minimal $\omega_{0}=\omega_{\text {min }}$ necessary for the dipole to reverse its orientation during the motion.
Context answer:
\boxed{$2 \frac{q B}{m}$}


Question: (d) (1 pt) If the dipole starts with $\vec{v}_{0}=0$ and $\omega_{0}=\omega_{\text {min }}$ found in part (c), the trajectory of its center of mass has an asymptote. Find the distance $D$ from the origin to the asymptote.

Solution:
If dipole's trajectory has an asymptote, then its movement along the asymptote is uniform. Indeed, if there is a linear motion with acceleration, the dipole $\vec{p}$ should be always aligned with the direction of motion, thus, not rotating. and as we found in part (a), the absence of rotation can only be maintained if $\vec{v}=$ const and $\vec{v} \perp \vec{p}$.

The uniform linear motion requires $\omega=0$, and this happens in the limit when the orientation is reversed $\vec{p}_{1}=-\vec{p}_{0}$. According to (11), in the limit, the dipole is travelling with the speed $\vec{v}_{1}=p_{0} B \hat{\jmath} / m$. Thus the asymptote is parallel to $\mathrm{Y}$ axis: $x=D$ (for counter-clockwise initial rotation).

<img_4425>

If $\vec{R}_{+}$and $\vec{R}_{-}$are absolute positions of the charges, we can write equation for the angular momentum around the origin $L_{O}$ :

$$
\begin{aligned}
& \frac{d \vec{L}_{O}}{d t}=\vec{R}_{+} \times\left(q \dot{\vec{R}}_{+} \times \vec{B}\right)+\vec{R}_{-} \times\left(-q \dot{\vec{R}}_{-} \times \vec{B}\right)= \\
& -q \vec{B}\left(\vec{R}_{+} \cdot \dot{\vec{R}}_{+}-\vec{R}_{-} \cdot \dot{\vec{R}}_{-}\right)=-\frac{q \vec{B}}{2} \frac{d}{d t}\left(R_{+}^{2}-R_{-}^{2}\right)
\end{aligned}
$$

After integration, we find one more conservation law (conservation of the "generalized angular momentum"):

$$
\begin{aligned}
\vec{L}_{O}+\frac{q \vec{B}}{2}\left(R_{+}^{2}-R_{-}^{2}\right)=\vec{L}_{O}+ & \frac{q \vec{B}}{2}\left(\left(\vec{R}_{+}+\vec{R}_{-}\right) \cdot\left(\vec{R}_{+}-\vec{R}_{-}\right)\right) \\
& =\vec{L}_{O}+\vec{B}(\vec{R} \cdot \vec{p})=\mathrm{const},
\end{aligned}
$$

where $\vec{R}=\frac{1}{2}\left(\vec{R}_{+}+\vec{R}_{-}\right)$is the position of center of mass. We also used the fact that $q\left(\vec{R}_{+}-\vec{R}_{-}\right)=2 q \vec{r}=\vec{p}$.

Initially, centre of mass coincides with origin $\left(\vec{R}_{0}=0\right)$ :

$$
L_{O}(0)=I \omega_{0}=2 m \frac{d^{2}}{4} 2 \frac{q B}{m}=q B d^{2}
$$

At asymptote, the dipole has reversed direction $\vec{p}_{1}=-\vec{p}_{0}$ and charges are travelling along parallel lines $x=D \pm r$ with the velocity $\vec{v}_{1}$ :

$$
\begin{aligned}
L_{O}(\infty)+ & B\left(\vec{R}_{1} \cdot \vec{p}_{1}\right)=m(D-r) v_{1}+m(D+r) v_{1}-B D p_{0} \\
& =2 m D \frac{p_{0} B}{m}-B D p_{0}=B D p_{0}=B D q d
\end{aligned}
$$

Since (13) equals (14), we conclude that $D=d$.



We can arrive to the same conclusion differently. Notice that we are interested in the $x$ coordinate of $C$ at infinity:

$$
D=x_{\infty}=\int_{0}^{\infty} v_{x} d t=\frac{q B d}{2 m} \int_{0}^{\infty} \sin \theta d t
$$

From (12), we can express $\sin \theta$ :

$$
\begin{aligned}
\int_{0}^{\infty} \sin \theta d t= & -\frac{m^{2}}{q^{2} B^{2}} \int_{0}^{\infty} \ddot{\theta} d t= \\
& -\frac{m^{2}}{q^{2} B^{2}}\left(\dot{\theta}_{1}-\dot{\theta}_{0}\right)=\frac{m^{2}}{q^{2} B^{2}} \omega_{\min }=\frac{2 m}{q B} .
\end{aligned}
$$

Finally,

$$
D=\frac{q B d}{2 m} \frac{2 m}{q B}=d
$$

Note. If initial rotation is clockwise $\left(\omega_{0}<0\right)$, the asymptote has an equation $x=-D$, but the distance to the origin remains the same.

Final answer: $D=d$