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 $X Y$.

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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 $(x_{c}, y_{c})=( \pm R_{c}, 0)$


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.

Solution:
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:

$$
\vec{v}_{1}=\vec{v}_{0}+\frac{\left(\vec{p}_{1}-\vec{p}_{0}\right) \times \vec{B}}{2 m}=-\frac{\vec{p}_{0} \times \vec{B}}{m}
$$



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:

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

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:

$$
\omega_{\min }=\dot{\theta}_{0}=\sqrt{4 \frac{g}{L}}=2 \frac{q B}{m}
$$

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

Final answer: $2 \frac{q B}{m}$