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APhO 2001, Теория — Задача 2

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Всички материали от APhO 2001

Theoretical Question 2 — Motion of an Electric Dipole in a Magnetic Field

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APHO II 2001 Theoretical Question 2 p. 1 / 2

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APHO II 2001 Theoretical Question 2 p. 2 / 2

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APHO II 2001 Theoretical Question 2 p. 1 / 4

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APHO II 2001 Theoretical Question 2 p. 2 / 4

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APHO II 2001 Theoretical Question 2 p. 3 / 4

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APHO II 2001 Theoretical Question 2 p. 4 / 4

Условие

Motion of an Electric Dipole in a Magnetic Field

In the presence of a constant and uniform magnetic field $\vec{B}$ , the translational motion of a system of electric charges is coupled to its rotational motion. As a result, the conservation laws for the momentum and the component of the angular momentum along the direction of $\vec{B}$ are modified from the usual form. This is illustrated in this problem by considering the motion of an electric dipole made of two particles of equal mass $m$ and carrying charges $q$ and $-q$ respectively ( $q > 0$ ). The two particles are connected by a rigid insulating rod of length $\ell$ , the mass of which can be neglected. Let $\vec{r}_1$ be the position vector of the particle with charge $q$ , $\vec{r}_2$ that of the other particle and $\vec{\ell} = \vec{r}_1 - \vec{r}_2$ . Denote by $\vec{\omega}$ the angular velocity of the rotation around the center of mass of the dipole. Denote by $\vec{r}_{CM}$ and $\vec{v}_{CM}$ the position and the velocity vectors of the center of mass respectively. Relativistic effects and effects of electromagnetic radiation can be neglected.

Note that the magnetic force acting on a particle of charge $q$ and velocity $\vec{v}$ is $q\vec{v} \times \vec{B}$ , where the cross product of two vectors $\vec{A}_1 \times \vec{A}_2$ is defined, in terms of the x, y, z, components of the vectors, by

(\vec{A}_1 \times \vec{A}_2)_x = (\vec{A}_1)_y (\vec{A}_2)_z - (\vec{A}_1)_z (\vec{A}_2)_y

(\vec{A}_1 \times \vec{A}_2)_y = (\vec{A}_1)_z (\vec{A}_2)_x - (\vec{A}_1)_x (\vec{A}_2)_z

(\vec{A}_1 \times \vec{A}_2)_z = (\vec{A}_1)_x (\vec{A}_2)_y - (\vec{A}_1)_y (\vec{A}_2)_x.

(1) Conservation Laws

(a) Write down the equations of motion for the center of mass of the dipole and for the rotation around the center of mass by computing the total force and the total torque with respect to the center of mass acting on the dipole.

(b) From the equation of motion for the center of mass, obtain the modified form of the conservation law for the total momentum. Denote the corresponding modified conserved quantity by $\vec{P}$ . Write down an expression in terms of $\vec{v}_{CM}$ and $\vec{\omega}$ for the conserved energy E.

(c) The angular momentum consists of two parts. One part is due to the motion of the center of mass and the other is due to rotation around the center of mass. From the modified form of the conservation law for the total momentum and the equation of motion of the rotation around the center of mass, prove that the quantity $J$ as defined by

J = (\vec{r}_{CM} \times \vec{P} + I\vec{\omega}) \cdot \hat{B}

is conserved.

Note that

\vec{A}_1 \times \vec{A}_2 = -\vec{A}_2 \times \vec{A}_1

\vec{A}_1 \cdot (\vec{A}_2 \times \vec{A}_3) = (\vec{A}_1 \times \vec{A}_2) \cdot \vec{A}_3

\vec{A}_1 \times (\vec{A}_2 \times \vec{A}_3) = (\vec{A}_1 \cdot \vec{A}_3)\vec{A}_2 - (\vec{A}_1 \cdot \vec{A}_2)\vec{A}_3

for any three vectors $\vec{A}_1$ , $\vec{A}_2$ and $\vec{A}_3$ . Repeated application of the above first two formulas may be useful in deriving the conservation law in question.

In the following, let $\vec{B}$ be in the z-direction.

(2) Motion in a Plane Perpendicular to $\vec{B}$

Suppose initially the center of mass of the dipole is at rest at the origin, $\vec{\ell}$ points in the x-direction and the initial angular velocity of the dipole is $\omega_0 \hat{z}$ ( $\hat{z}$ is the unit vector in the z-direction).

(a) If the magnitude of $\omega_0$ is smaller than a critical value $\omega_c$ , the dipole will not make a full turn with respect to its center of mass. Find $\omega_c$ .

(b) For a general $\omega_0 > 0$ , what is the maximum distance $d_m$ in the x-direction that the center of mass can reach?

Решение

(1) Conservation Laws

(1a) $\vec{r}_{CM} = \frac{1}{2}(\vec{r}_1 + \vec{r}_2)$ , $\vec{v}_{CM} = \frac{1}{2}(\vec{v}_1 + \vec{v}_2)$ , $\vec{\ell} = \vec{r}_1 - \vec{r}_2$ , $\vec{u} = \dot{\vec{\ell}} = \vec{v}_1 - \vec{v}_2$

Total force $\vec{F}$ on the dipole is

\vec{F} = \vec{F}_1 + \vec{F}_2 = q(\vec{E} + \vec{v}_1 \times \vec{B}) + (-q)(\vec{E} + \vec{v}_2 \times \vec{B}) = q(\vec{v}_1 - \vec{v}_2) \times \vec{B}

= q\dot{\vec{\ell}} \times \vec{B}

so that

M\dot{\vec{v}}_{CM} = q\dot{\vec{\ell}} \times \vec{B} \qquad (M = 2m) \qquad (1)

Computing the torque for rotation around the center of mass, we obtain

I\dot{\vec{\omega}} = (\frac{\vec{\ell}}{2}) \times (q\vec{v}_1 \times \vec{B}) + (\frac{-\vec{\ell}}{2}) \times (-q\vec{v}_2 \times \vec{B})

= q\vec{\ell} \times (\vec{v}_{CM} \times \vec{B}) \qquad (2)

where

I = \frac{1}{2} m\ell^2 \qquad (3)

(1b) From eq.(1), we obtain the conservation law for the momentum:

\dot{\vec{P}} = 0, \quad \vec{P} = M\vec{v}_{CM} - q\vec{\ell} \times \vec{B} \qquad (4)

From eq.(1) and eq.(2), one obtains the conservation law for the energy.

\dot{E} = 0, \quad E = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} I\omega^2 \qquad (5)

(1c) Using eq.(4) and eq.(2),

\frac{d}{dt}(\vec{r}_{CM} \times \vec{P}) \cdot \hat{B} = (\vec{v}_{CM} \times \vec{P}) \cdot \hat{B} = -q\vec{v}_{CM} \times (\vec{\ell} \times \vec{B}) \cdot \hat{B}

= q(\vec{\ell} \times \vec{B}) \times \vec{v}_{CM} \cdot \hat{B} = q(\vec{\ell} \times \vec{B}) \cdot (\vec{v}_{CM} \times \hat{B})

= q\vec{\ell} \cdot (\vec{B} \times (\vec{v}_{CM} \times \hat{B})) = -q\vec{\ell} \cdot ((\vec{v}_{CM} \times \vec{B}) \times \hat{B})

= -q\vec{\ell} \times (\vec{v}_{CM} \times \vec{B}) \cdot \hat{B}

= -I\dot{\vec{\omega}} \cdot \hat{B}

we obtain the conservation law

\dot{J} = 0 \qquad J = (\vec{r}_{CM} \times \vec{P} + I\vec{\omega}) \cdot \hat{B} \qquad (6)

for the component of the angular momentum along the direction of $\vec{B}$ .

(2) Motion in a Plane Perpendicular to $\vec{B}$

(2a) Write

\vec{\ell} = \ell\{\cos\varphi(t)\hat{x} + \sin\varphi(t)\hat{y}\}, \quad \varphi(0) = 0, \quad \dot{\varphi}(0) = \omega_0 \qquad (7)

Note that

\vec{\omega} = \dot{\varphi}\hat{z} \qquad (8)

From eq.(4), we have

M\vec{v}_{CM} = \vec{P} + q\ell B(\sin\varphi\,\hat{x} - \cos\varphi\,\hat{y}) \qquad (9)

At $t = 0$ , we have $v_{CM} = 0$ , $\varphi = 0$ so that

\vec{P} = q\ell B\hat{y} \qquad (10)

Hence from eqs.(9) and (10) we have

\dot{x}_{CM} = \left(\frac{q\ell B}{M}\right)\sin\varphi, \quad \dot{y}_{CM} = \left(\frac{q\ell B}{M}\right)(1 - \cos\varphi) \qquad (11)

From conservation of energy, i.e. Eq.(5), we have

\frac{1}{2}I\dot{\varphi}^2 + \frac{(q\ell B)^2}{M}(1 - \cos\varphi) = \frac{1}{2}I\omega_0^2

\therefore \quad \dot{\varphi}^2 + \frac{1}{2}\omega_c^2(1 - \cos\varphi) = \omega_0^2 \qquad (12)

where

\omega_c^2 = \frac{4(q\ell B)^2}{MI} = \left(\frac{2qB}{m}\right)^2 \qquad (13)

In order to make a full turn, $\dot{\varphi}$ can not become zero so that

\omega_0^2 > \omega_c^2 \implies |\omega_0| > \omega_c = \frac{2qB}{m} \qquad (14)

(2b) From Eq.(6), we have

x_{CM}P + I\omega = J \qquad (15)

where $P$ is the magnitude of $\vec{P}$ .

At $t = 0$ , we have $J = I\omega_0$ so that

x_{CM}P + I\omega = I\omega_0 \qquad (16)

From eq.(12), one can see that $\omega_0^2 \geq \omega^2$ so that $x_{CM} \geq 0$ . Thus $x_{CM}$ reaches a maximum $d_m$ when $\omega$ takes its minimum value.

When $\omega_0 < \omega_c$ , the minimum value of $\omega$ is $-\omega_0$ so that

d_m = \frac{2I}{P}\omega_0 = \left(\frac{m\omega_0}{qB}\right)\ell, \quad \omega_0 < \omega_c \qquad (17)

When $\omega_0 > \omega_c$ , the minimum value of $\omega$ is $\sqrt{\omega_0^2 - \omega_c^2}$ so that

d_m = \left(\frac{I}{P}\right)\left(\omega_0 - \sqrt{\omega_0^2 - \omega_c^2}\right) = \frac{m}{2qB}\left(\omega_0 - \sqrt{\omega_0^2 - \omega_c^2}\right)\ell, \quad \omega_0 > \omega_c \qquad (18)

When $\omega_0 = \omega_c$ , $\omega^2 = \frac{1}{2}\omega_c^2(1 + \cos\varphi) = \omega_c^2\cos^2\frac{\phi}{2}$

\therefore \quad \dot{\phi} = \omega_c\cos\frac{\phi}{2}

When $\varphi$ is close to $\pi$ , let $\varphi = \pi - 2\varepsilon$ then

\dot{\varepsilon} = -\frac{1}{2}\omega_c\sin\varepsilon \approx -\frac{1}{2}\omega_c\varepsilon

\therefore \quad \varepsilon \sim e^{-\omega_c t/2}

so that it will take $t \to \infty$ for $\varepsilon \to 0$ , i.e. for $\varphi$ to reach $\pi$ . Hence

d_m = \left(\frac{I}{P}\right)\omega_c = \left(\frac{m\omega_c}{2qB}\right)\ell, \quad \omega_0 = \omega_c \qquad (19)

(2c) Tension on the rod comes from three sources:

(i) Coulomb force

= \frac{1}{4\pi\varepsilon_0}\frac{q^2}{\ell^2} \qquad (20)

Positive value means compression on the rod.

(ii) Centrifugal force due to rotation of the rod

= -\frac{1}{2}m\omega^2\ell \qquad (21)

(iii) Magnetic force on the particles due to the motion of the center of the mass

= q\vec{v}_{CM} \times \vec{B} \cdot (-\hat{\ell}) = q\vec{v}_{CM} \cdot \hat{\ell} \times \vec{B}

Taking the square of both sides of eq.(4) and using the initial condition for the value of $P^2$ , we obtain

\frac{1}{2}Mv_{CM}^2 = q\ell\vec{v}_{CM} \cdot \hat{\ell} \times \vec{B} = \frac{1}{2}I(\omega_0^2 - \omega^2) \qquad (22)

Combining the three forces, we have

\text{tension on the rod} = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{\ell^2} - \frac{1}{2}m\ell\omega^2 + \frac{1}{4}m\ell(\omega_0^2 - \omega^2) \qquad (23)

A positive value means compression on the rod.

Теми и трудност

Трудност: неоценена.

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Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf