EuPhO 2022, Теория — Задача 3. Dipole in a magnetic field

• (a) ω₀ = 0 and v⃗₀ ∥ Y (perpendicular to the dipole).

• (b) $\vec v_0 = -\frac{qdB}{2m}\hat\jmath, \quad R_c = \frac{qBd}{2m|\omega_0|}, \quad (x_c, y_c) = (\pm R_c, 0)$

  The “+” sign corresponds to ω₀ > 0 (counter-clockwise rotation), the “−” sign to clockwise rotation.

• (c) $\omega_{\mathrm{min}} = \frac{2qB}{m}$

• (d) $D = d$

Part (a): Uniform linear motion

Lorentz forces acting on the charges:

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:

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 $C$:

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.

In summary, the dipole will move uniformly along $Y$ if, and only if, $\vec v_0\parallel Y$ and $\omega_0 = 0$.

Part (b): Circular motion

The net force can be calculated as:

where $\vec p$ is a dipole moment ($|\vec p| = qd = 2qr$ 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 $\vec F\parallel\vec p$, 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:

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

i.e. the magnitude of velocity is:

and the radius of the orbit is:

The coordinates of the center of the circle are:

where the “$+$” sign corresponds to $\omega_0 > 0$, i.e. counter-clockwise rotation, and the “$-$” sign —to clockwise rotation. In either case, the initial velocity should point to the negative $Y$ direction:

Part (c): Reversal of the dipole

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

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

From Newton’s second law:

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

Thus, if $\vec p$ has reversed its direction from $\vec p_0$ to $\vec p_1 = -\vec p_0$, then the velocity:

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

Here, $I = 2\times m(d/2)^2 = md^2/2$ 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,

Alternatively, we can introduce $\theta$ to be the angle between the dipole moment and the axis $X$ ($\theta_0 = 0$) and rewrite the equations of translational motion in coordinates using $\omega = \dot\theta$:

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

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

This is the equation of a mathematical pendulum of length $L$ in gravitational field $g = L(qB/m)^2$. 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 $K = \frac{1}{2}mL^2\dot\theta_0^2$ will be transferred to the potential energy $U = 2mgL$, from which we find:

Note. Due to symmetry, both clockwise and counter-clockwise initial rotation with absolute value of $|\omega_0|$ will work.

Part (d): 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 $\vec v = \mathrm{const}$ 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 $\vec v_1 = p_0B\hat\jmath/m$. 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 $L_O$:

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

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

Initially, centre of mass coincides with origin ($\vec R_0 = 0$):

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 $\vec v_1$:

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:

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

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.

Marking scheme

Only one of the grading tables should be used for part (c), the one which results in a higher score.

Alternative approach: pendulum analogy