IdPhO 2020, Теория — Задача 1C. Part C. Rod in magnetic field

• C.1 $z_{\max} = \frac{\sqrt{2}mv}{qB}\sqrt{\sqrt{1 + \left(\frac{qBR}{mv}\right)^2} - 1},\quad a = \sqrt{\left(\frac{v^2}{R}\right)^2 + \left(\frac{qvBz_{\max}}{mR}\right)^2}$

The equations of motion of the centre of mass of balls is:

where $\vec{v}_C$ is the velocity of the centre of mass, $\vec{v}_1$ and $\vec{v}_2$ are velocities of balls. Notice that $v_C = 0$ at the start of motion then $v_C$ remains equal to $0$ always.

The equation for the rotation of the ball about the centre of mass is:

Now let's use the formula of a triple vector product:

So balls move around the sphere with the center in the middle of rod. It means that $\vec{R} \perp \vec{v}$ then $\vec{R} \cdot \vec{v} = 0$. Hence

Firstly, $\vec{B}$ directs along $z$-axis, then $\left(\vec{R} \cdot \vec{B}\right) = Bz$, where $z$ is a coordinate of the ball. Secondly, let look at projection of vector equation onto $z$-axis:

Multiplying to $dt$ and integrating gives us this equation:

where $C = 0$ because $L_z = 0$ at the start of the motion when $z = 0$.

When $z$ is maximum $v_z = \dot{z} = 0$. It means, that $L = L_z = mur$, where $r$ is a distance between the ball and $z$-axis, $u$ is the velocity of the ball at this moment. Recall that the ball is moving around the sphere then $r^2 + z^2 = R^2$.

The Lorenz force and the contact ball-rod force don't do the work. It means the kinetic energy of the system is constant. Together with $\vec{v}_C = 0$, it gives that velocities of the balls don't change i.e. $u = v$.

To summarize, $L_z = mv\sqrt{R^2 - z_{\max}^2}$.

Let's combine the results of the two last blocks as a biquadratic equation for $z_{\max}$:

Solution is:

This result can be obtained in a faster way by those who are familiar with the generalized momentum of a charged particle in a magnetic field,

where $\vec{A}$ is the vector potential which is defined by the condition $\mathrm{curl}\vec{A} = \vec{B}$. We can choose $\vec{A} = \frac{1}{2}\hat{\tau}Br$, where $r$ denotes the distance from the $z$-axis, and $\hat{\tau}$ stands for the unit vector perpendicular both to the radius vector and to the $z$-axis. This is a vector field obeying cylindrical symmetry around the $z$-axis and because of that, the corresponding generalized momentum is conserved,

here $v_\perp$ denotes the velocity component perpendicular to the $z$-axis. Noting that at the topmost position of the ball's trajectory, $v_\perp = v$ due to the fact that the speed is constant and horizontal (i.e. in $x-y$-plane), we obtain a quadratic equation for $\rho = \frac{r}{R}$:

We select the positive root $\rho = \sqrt{\kappa^2 + 1} - \kappa$. Now we can write the final answer as

which coincides with the previous result.

Consider three orthogonal unit vectors: 1-st is along with the velocity, 2-nd along with direction from ball to the middle of the rod, and 3-rd is perpendicular to the other two.

As we found earlier that $v = \mathrm{const}$, so $a_1 = 0$. Further, $a_2$ coincides with $v^2/R$ because ball is moving around sphere. Finally, the projection of 2-nd Newton law onto the 3-rd axis is: $ma_3 = qvB\frac{z_{\max}}{R}$. Hence