(a)  ^2.00 Find $\omega_{0}$ and the direction of $\vec{v}_{0}$, so that the center of mass will move with the constant velocity $\vec{v}=\vec{v}_{0}$ ?

Uniform linear motion

Lorentz forces acting on the charges:

$$
\begin{gathered}
\vec{F}_{+}=q \vec{v}_{+} \times \vec{B}=q(\vec{v}+\vec{\omega} \times \vec{r}) \times \vec{B}, \\
\vec{F}_{-}=(-q) \vec{v}_{-} \times \vec{B}=(-q)(\vec{v}-\vec{\omega} \times \vec{r}) \times \vec{B},
\end{gathered}
$$where $\vec{r}$ is a vector from the center of mass to the position of the positive charge.

According to Newton's first law, the center-of-mass $C$ of the dipole will move with constant velocity provided that the net force:

\begin{equation*}
\vec{F}=\vec{F}_{+}+\vec{F}_{-}=q\left(\vec{v}_{+}-\vec{v}_{-}\right) \times \vec{B}, \tag{8}
\end{equation*}acting on the dipole, is zero. Since $\vec{v}_{+}, \vec{v}_{-}$and $\vec{B}$ are perpendicular, we require $\vec{v}_{+}=\vec{v}_{-}$. It means that dipole does not rotate: $\omega=\omega_{0}=0$.

The pure translation, however, is possible if the pair of forces $\vec{F}_{+}, \vec{F}_{-}$, has zero torque about $C$ :

\begin{align*}
& \vec{\tau}=\vec{r} \times \vec{F}_{+}-\vec{r} \times \vec{F}_{-}=2 q \vec{r} \times(\vec{v} \times \vec{B})= \\
& \quad 2 q(\vec{v}(\vec{r} \cdot \vec{B})-\vec{B}(\vec{r} \cdot \vec{v}))=-2 q \vec{B}(\vec{r} \cdot \vec{v}) \tag{9}
\end{align*}

We conclude that scalar product is zero only when $\vec{v} \perp \vec{r}$, i.e. the initial velocity should be parallel to $Y$ direction.

(b)  ^3.00 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.

Circular motion

The net force can be calculated as:

\begin{align*}
\vec{F}= & \vec{F}_{+}+\vec{F}_{-}=2 q(\vec{\omega} \times \vec{r}) \times \vec{B}= \\
& -2 q(\vec{\omega}(\vec{B} \cdot \vec{r})-\vec{r}(\vec{B} \cdot \vec{\omega}))=2 q B \omega \vec{r}=B \omega \vec{p}, \tag{10}
\end{align*}

where $\vec{p}$ is a dipole moment $(|\vec{p}|=q d=2 q r$ and the direction aligns with $\vec{r}$ ).

When $C$ orbits a circle, $\vec{F}$ acts as a centripetal force, i.e. it points to the center of the circle. Since $\vec{F} \| \vec{p}$, the dipole is always in line with the center of the orbit. Therefore, the orbital angular velocity of $C$ is equal to the angular velocity of rotation of the dipole about $C$.

The magnitude of the orbital velocity is:

$$
v_{0}=\left|\omega_{0}\right| R_{c}
$$

From Newton's second law, and accounting that the total mass of the dipole is $2 m$ :

$$
\frac{2 m v_{0}^{2}}{R_{c}}=\frac{p B v_{0}}{R_{c}},
$$

i.e. the magnitude of velocity is:

$$
v_{0}=\frac{p B}{2 m}=\frac{q B d}{2 m}
$$

and the radius of the orbit is:

(c)  ^4.00 Given $\vec{v}_{0}=0$, find the minimal $\omega_{0}=\omega_{\text {min }}$ necessary for the dipole to reverse its orientation during the motion.

Reversal of the dipole

In (10) we have shown that the net force:

$$
\vec{F}=2 q(\vec{\omega} \times \vec{r}) \times \vec{B}=(\vec{\omega} \times \vec{p}) \times \vec{B} .
$$

Since the dipole moment $\vec{p}$ rotates with angular velocity $\vec{\omega}$, its time derivative:

$$
\frac{d \vec{p}}{d t}=\vec{\omega} \times \vec{p} .
$$

From Newton's second law:

$$
2 m \frac{d \vec{v}}{d t}=\vec{F}=\frac{d \vec{p}}{d t} \times \vec{B} .
$$

By integrating the equation, we arrive at an additional conservation law in the system (conservation of the so called "generalized momentum"):

$$
2 m \vec{v}-\vec{p} \times \vec{B}=\mathrm{const}
$$

Thus, if $\vec{p}$ has reversed its direction from $\vec{p}_{0}$ to $\vec{p}_{1}=-\vec{p}_{0}$, then the velocity:

\begin{equation*}
\vec{v}_{1}=\vec{v}_{0}+\frac{\left(\vec{p}_{1}-\vec{p}_{0}\right) \times \vec{B}}{2 m}=-\frac{\vec{p}_{0} \times \vec{B}}{m} . \tag{11}
\end{equation*}

Since the magnetic field does not perform work on moving electric charges, the kinetic energy of the dipole is conserved:

$$
\frac{I}{2} \omega_{0}^{2}=\frac{I}{2} \omega_{1}^{2}+\frac{2 m}{2} v_{1}^{2},
$$

Here, $I=2 \times m(d / 2)^{2}=m d^{2} / 2$ is the moment of inertia of the dipole with respect to its center-of-mass. Since $v_{1}$ doesn't depend on angular velocities, $\omega_{0}$ is minimal when $\omega_{1}=0$. Finally,

$$
\omega_{\min }=v_{1} \sqrt{\frac{2 m}{I}}=\frac{p_{0} B}{m} \sqrt{\frac{4}{d^{2}}}=\frac{2 q B}{m}
$$

Alternatively, we can introduce $\theta$ to be the angle between the dipole moment and the axis $X\left(\theta_{0}=0\right)$ and rewrite the equations of translational motion in coordinates using $\omega=\dot{\theta}$ :

$$
\dot{v}_{x}=\dot{\theta} \frac{q B d}{2 m} \cos \theta, \quad \dot{v}_{y}=\dot{\theta} \frac{q B d}{2 m} \sin \theta .
$$

By integrating these equations, given zero initial velocity, we find how velocity depends on $\theta$ :

$$
v_{x}=\frac{q B d}{2 m} \sin \theta, \quad v_{y}=\frac{q B d}{2 m}(1-\cos \theta) .
$$

Using the expression (9) for the torque, we can write the equation of rotational motion as:

\begin{gather*}
I \ddot{\theta}=\tau=-2 q B\left(r_{x} v_{x}+r_{y} v_{y}\right)=-\frac{q^{2} B^{2} d^{2}}{2 m} \sin \theta \\
\ddot{\theta}+\frac{q^{2} B^{2}}{m^{2}} \sin \theta=0 \tag{12}
\end{gather*}

This is the equation of a mathematical pendulum of length $L$ in gravitational field $g=L(q B / m)^{2}$. And the equivalent question becomes what is the minimal push $\dot{\theta}_{0}$ required in the bottom position for the pendulum to reach the top position. Kinetic energy of the pendulum $K=\frac{1}{2} m L^{2} \dot{\theta}_{0}^{2}$ will be transfered to the potential energy $U=2 m g L$, from which we find:

Note. Due to symmetry, both clockwise and counterclockwise initial rotation with absolute value of $\left|\omega_{0}\right|$ will work.

(d)  ^1.00 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.

Trajectory asymptote

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 Y axis: $x=D$ (for counter-clockwise initial rotation).

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"):

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

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 ( $\vec{R}_{0}=0$ ):

\begin{equation*}
L_{O}(0)=I \omega_{0}=2 m \frac{d^{2}}{4} 2 \frac{q B}{m}=q B d^{2} . \tag{13}
\end{equation*}

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{align*}
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 \tag{14}
\end{align*}

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,

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