Question

A dipole consists of two equal and opposite charges separated by a small distance. One such a dipole with charges +q and -q each, and mass m each connected by a massless rigid rod of length l, is rotating with angular speed $\omega_0$ about -ve z-axis in XY plane with its centre of mass at origin with zero initial velocity. There exists uniform magnetic field B in the region along -ve z-direction. Select correct option(s).

• A. The maximum speed of the centre of mass of the dipole during the motion is $\frac{BqL}{m}$.
• B. If the magnitude of $\omega_0$ is greater than 2qB/m, the dipole will not stop rotating.
• C. If the angular velocity becomes zero, speed of center of mass is maximum at that time.
• D. The maximum speed of the centre of mass of the dipole during the motion is $\frac{\omega_0 L}{2}$.

Answer 1

Answer: (C), (D)

C,D

Answer 2

The problem describes a dipole consisting of two charges, +q and -q, each of mass m, connected by a massless rigid rod of length L. The dipole rotates with initial angular speed $\omega_0$ about the -ve z-axis in the XY plane, with its center of mass (CM) initially at the origin and having zero initial velocity. A uniform magnetic field B exists along the -ve z-direction.

Let's analyze the forces and torques acting on the dipole. Let $\vec{r}'_+$ be the position vector of charge +q relative to the CM, so $|\vec{r}'_+| = L/2$. Let $\vec{r}'_-$ be the position vector of charge -q relative to the CM, so $|\vec{r}'_-| = L/2$. Note that $\vec{r}'_- = -\vec{r}'_+$.

Let $\vec{V}_{CM}$ be the velocity of the center of mass and $\vec{\omega}$ be the instantaneous angular velocity of the dipole. The velocity of charge +q is $\vec{v}_+ = \vec{V}_{CM} + \vec{\omega} \times \vec{r}'_+$. The velocity of charge -q is $\vec{v}_- = \vec{V}_{CM} + \vec{\omega} \times \vec{r}'_- = \vec{V}_{CM} - \vec{\omega} \times \vec{r}'_+$.

The magnetic force on charge +q is $\vec{F}_+ = q(\vec{v}_+ \times \vec{B})$. The magnetic force on charge -q is $\vec{F}_- = -q(\vec{v}_- \times \vec{B})$.

1. Net Force on the Dipole (on CM): The net force on the dipole is $\vec{F}_{net} = \vec{F}_+ + \vec{F}_-$. $\vec{F}_{net} = q(\vec{V}_{CM} + \vec{\omega} \times \vec{r}'_+) \times \vec{B} - q(\vec{V}_{CM} - \vec{\omega} \times \vec{r}'_+) \times \vec{B}$ $\vec{F}_{net} = q(\vec{V}_{CM} \times \vec{B}) + q((\vec{\omega} \times \vec{r}'_+) \times \vec{B}) - q(\vec{V}_{CM} \times \vec{B}) + q((\vec{\omega} \times \vec{r}'_+) \times \vec{B})$ $\vec{F}_{net} = 2q ((\vec{\omega} \times \vec{r}'_+) \times \vec{B})$.

Given $\vec{\omega}$ is along -z, so $\vec{\omega} = -\omega \hat{k}$. Given $\vec{B}$ is along -z, so $\vec{B} = -B \hat{k}$. Let $\vec{r}'_+$ be along the x-axis for an instant, so $\vec{r}'_+ = (L/2)\hat{i}$. Then $\vec{\omega} \times \vec{r}'_+ = (-\omega \hat{k}) \times (L/2)\hat{i} = -\omega(L/2)(\hat{k} \times \hat{i}) = -\omega(L/2)\hat{j}$. Now, $((\vec{\omega} \times \vec{r}'_+) \times \vec{B}) = (-\omega(L/2)\hat{j}) \times (-B\hat{k}) = \omega B (L/2)(\hat{j} \times \hat{k}) = \omega B (L/2)\hat{i}$. So, $\vec{F}_{net} = 2q (\omega B (L/2)\hat{i}) = qBL\omega \hat{i}$. In general, if $\vec{r}'_+$ is the vector from CM to +q, then $\vec{F}_{net}$ is always directed along $\vec{r}'_+$ (radially outwards from the CM). The magnitude of the net force is $F_{net} = qBL\omega$. The total mass of the dipole is $M = 2m$. The acceleration of the CM is $\vec{a}_{CM} = \frac{\vec{F}_{net}}{M} = \frac{qBL\omega}{2m} \hat{r}'_+$.

2. Net Torque on the Dipole about CM: The net torque about the CM is $\vec{\tau}_{CM} = \vec{r}'_+ \times \vec{F}_+ + \vec{r}'_- \times \vec{F}_-$. $\vec{\tau}_{CM} = \vec{r}'_+ \times [q(\vec{V}_{CM} \times \vec{B}) + q((\vec{\omega} \times \vec{r}'_+) \times \vec{B})] + (-\vec{r}'_+) \times [-q(\vec{V}_{CM} \times \vec{B}) + q((\vec{\omega} \times \vec{r}'_+) \times \vec{B})]$ $\vec{\tau}_{CM} = q \vec{r}'_+ \times (\vec{V}_{CM} \times \vec{B}) + q \vec{r}'_+ \times ((\vec{\omega} \times \vec{r}'_+) \times \vec{B}) + q \vec{r}'_+ \times (\vec{V}_{CM} \times \vec{B}) - q \vec{r}'_+ \times ((\vec{\omega} \times \vec{r}'_+) \times \vec{B})$ $\vec{\tau}_{CM} = 2q [\vec{r}'_+ \times (\vec{V}_{CM} \times \vec{B})]$.

Let $\vec{r}'_+ = (L/2)(\cos\phi \hat{i} + \sin\phi \hat{j})$ where $\phi$ is the angle of the dipole with the x-axis. Let $\vec{V}_{CM} = V_{CMx} \hat{i} + V_{CMy} \hat{j}$. $\vec{V}_{CM} \times \vec{B} = (V_{CMx} \hat{i} + V_{CMy} \hat{j}) \times (-B \hat{k}) = -B V_{CMx}(\hat{i} \times \hat{k}) - B V_{CMy}(\hat{j} \times \hat{k})$ $= -B V_{CMx}(-\hat{j}) - B V_{CMy}(\hat{i}) = B V_{CMx}\hat{j} - B V_{CMy}\hat{i}$.

$\vec{\tau}_{CM} = 2q (L/2)(\cos\phi \hat{i} + \sin\phi \hat{j}) \times (B V_{CMx}\hat{j} - B V_{CMy}\hat{i})$ $\vec{\tau}_{CM} = qBL [(\cos\phi \hat{i} + \sin\phi \hat{j}) \times (V_{CMx}\hat{j} - V_{CMy}\hat{i})]$ $\vec{\tau}_{CM} = qBL [\cos\phi V_{CMx}(\hat{i} \times \hat{j}) - \cos\phi V_{CMy}(\hat{i} \times \hat{i}) + \sin\phi V_{CMx}(\hat{j} \times \hat{j}) - \sin\phi V_{CMy}(\hat{j} \times \hat{i})]$ $\vec{\tau}_{CM} = qBL [\cos\phi V_{CMx} \hat{k} + \sin\phi V_{CMy} \hat{k}]$ $\vec{\tau}_{CM} = qBL (V_{CMx}\cos\phi + V_{CMy}\sin\phi) \hat{k}$. The term $(V_{CMx}\cos\phi + V_{CMy}\sin\phi)$ is the component of $\vec{V}_{CM}$ along the dipole axis $\hat{r}'_+$. Let's call this $V_{CM, \parallel}$. So, $\vec{\tau}_{CM} = qBL V_{CM, \parallel} \hat{k}$. The angular velocity is $\vec{\omega} = -\omega \hat{k}$. The moment of inertia about the CM for rotation in XY plane is $I = m(L/2)^2 + m(L/2)^2 = mL^2/2$. The equation of rotational motion is $I \frac{d\vec{\omega}}{dt} = \vec{\tau}_{CM}$. $mL^2/2 \frac{d(-\omega)}{dt} \hat{k} = qBL V_{CM, \parallel} \hat{k}$. $-mL^2/2 \frac{d\omega}{dt} = qBL V_{CM, \parallel}$.

3. Energy Conservation: Magnetic forces do no work. Therefore, the total mechanical energy of the system is conserved. The total energy $E = K_{trans} + K_{rot} = \frac{1}{2} M V_{CM}^2 + \frac{1}{2} I \omega^2$. $E = \frac{1}{2} (2m) V_{CM}^2 + \frac{1}{2} (mL^2/2) \omega^2 = m V_{CM}^2 + \frac{1}{4} mL^2 \omega^2$. Initial state: $V_{CM,initial} = 0$, $\omega = \omega_0$. Initial energy $E_0 = m(0)^2 + \frac{1}{4} mL^2 \omega_0^2 = \frac{1}{4} mL^2 \omega_0^2$. So, throughout the motion: $m V_{CM}^2 + \frac{1}{4} mL^2 \omega^2 = \frac{1}{4} mL^2 \omega_0^2$.

Analysis of Options:

(A) The maximum speed of the centre of mass of the dipole during the motion is $\frac{BqL}{m}$. From the energy conservation equation, $m V_{CM}^2 = \frac{1}{4} mL^2 \omega_0^2 - \frac{1}{4} mL^2 \omega^2$. $V_{CM}^2 = \frac{1}{4} L^2 (\omega_0^2 - \omega^2)$. The maximum speed of the CM occurs when $\omega = 0$. $V_{CM,max}^2 = \frac{1}{4} L^2 \omega_0^2$. $V_{CM,max} = \frac{\omega_0 L}{2}$. This contradicts option (A). So (A) is incorrect.

(D) The maximum speed of the centre of mass of the dipole during the motion is $\frac{\omega_0 L}{2}$. As derived above, this is correct.

(C) If the angular velocity becomes zero, speed of center of mass is maximum at that time. From the energy conservation equation, $V_{CM}^2 = \frac{1}{4} L^2 (\omega_0^2 - \omega^2)$. This equation clearly shows that $V_{CM}$ is maximum when $\omega$ is minimum (i.e., $\omega=0$). So, (C) is correct.

(B) If the magnitude of $\omega_0$ is greater than 2qB/m, the dipole will not stop rotating. The angular speed $\omega$ becomes zero when $V_{CM}$ is maximum. The condition for $\omega$ to become zero is that the torque $\vec{\tau}_{CM}$ must act to decelerate the rotation. $\frac{d\omega}{dt} = -\frac{qBL V_{CM, \parallel}}{mL^2/2} = -\frac{2qB V_{CM, \parallel}}{mL}$. For $\omega$ to decrease, $V_{CM, \parallel}$ must be positive. The CM starts from rest, but the force on CM is $F_{net} = qBL\omega$. So the CM will accelerate. The acceleration of CM is $\vec{a}_{CM} = \frac{qBL\omega}{2m} \hat{r}'_+$. The CM moves in a trajectory that is not simple. Let's consider the initial state: $V_{CM}=0$. At $t=0$, $\tau_{CM} = qBL (V_{CMx}\cos\phi + V_{CMy}\sin\phi) \hat{k} = qBL (0) = 0$. This means initially, $\frac{d\omega}{dt}=0$. So $\omega$ remains $\omega_0$ for a small time. In this small time, $F_{net} = qBL\omega_0$. The CM accelerates and starts moving. $2m \frac{d\vec{V}_{CM}}{dt} = qBL\omega_0 \hat{r}'_+$. The CM starts moving in the direction of the dipole axis. So $\vec{V}_{CM}$ will be parallel to $\hat{r}'_+$. Thus, $V_{CM, \parallel} = V_{CM}$. Therefore, $\frac{d\omega}{dt} = -\frac{2qB V_{CM}}{mL}$. As $V_{CM}$ increases, $\omega$ decreases. As $\omega$ decreases, $F_{net}$ decreases, which means $a_{CM}$ decreases. The system will eventually reach a state where $\omega=0$ and $V_{CM}$ is maximum, provided the initial angular kinetic energy is completely converted to translational kinetic energy. The question is about whether the dipole will not stop rotating if $\omega_0$ is greater than $2qB/m$. This implies that there might be a condition where $\omega$ never reaches zero. If $\omega$ never reaches zero, then $V_{CM}$ can never reach its maximum value $\frac{\omega_0 L}{2}$. The total energy is conserved. The angular velocity will only become zero if $V_{CM}$ can reach $\frac{\omega_0 L}{2}$. The condition for stopping rotation is $\omega=0$. This is always possible as long as the CM can acquire kinetic energy. The magnetic force and torque are always present as long as there is rotation and/or CM motion. The equation for $\omega$ is $-mL^2/2 \frac{d\omega}{dt} = qBL V_{CM}$. If $\omega > 0$, then $V_{CM}$ increases, and $\omega$ decreases. This means that rotation will always tend to slow down as long as $V_{CM}$ is increasing. The energy conservation implies that $\omega$ can go to zero. The statement "will not stop rotating" would mean $\omega$ never becomes zero. This would happen if the energy conversion is somehow limited or if the torque becomes zero before $\omega$ reaches zero. The torque is $qBL V_{CM, \parallel}$. This torque depends on $V_{CM, \parallel}$. Since the CM starts from rest and accelerates along the dipole axis, $V_{CM, \parallel}$ will be non-zero and positive. So the torque will be non-zero and negative, continuously decreasing $\omega$. Thus, $\omega$ will always decrease until it reaches zero, at which point $V_{CM}$ is maximum. Therefore, the dipole will stop rotating. So, option (B) is incorrect.

Final check: (A) Incorrect. Max speed of CM is $\omega_0 L/2$. (B) Incorrect. The dipole will stop rotating when its rotational kinetic energy is fully converted to translational kinetic energy. (C) Correct. When $\omega=0$, $V_{CM}$ is maximum. (D) Correct. Max speed of CM is $\omega_0 L/2$.

The maximum speed of the center of mass is $V_{CM,max} = \frac{\omega_0 L}{2}$. At this point, $\omega=0$. The force on CM is $F_{net} = qBL\omega$. When $\omega=0$, $F_{net}=0$. The torque on the dipole is $\tau_{CM} = qBL V_{CM, \parallel}$. When $\omega=0$, $V_{CM}$ is maximum, and the CM is moving. The direction of $V_{CM}$ would be along the dipole axis. So $V_{CM, \parallel} = V_{CM,max}$. Thus, there is still a torque acting on the dipole at this point, which will cause it to start rotating in the opposite direction. The motion is oscillatory, converting rotational energy to translational energy and back.