Solution
System Analysis:
We have a dumbbell consisting of a rigid rod of length $l$. The ends contain unlike charges of equal modulus $q$. Let’s assign:
- Charge at end A: $+q$
- Charge at end B: $-q$
The system is in a uniform magnetic field $\vec{B} = B\hat{k}$.
The motion is a superposition of translation of the center of mass (velocity $\vec{v}_c$) and rotation about the center (angular velocity $\vec{\omega} = \omega\hat{k}$).
1. Calculation of Net Force
The velocities of the two ends are:
$$ \vec{v}_A = \vec{v}_c + \vec{\omega} \times \vec{r}_{A/C} $$ $$ \vec{v}_B = \vec{v}_c + \vec{\omega} \times \vec{r}_{B/C} $$The net magnetic force is the sum of forces on both particles:
$$ \vec{F}_{net} = q(\vec{v}_A \times \vec{B}) + (-q)(\vec{v}_B \times \vec{B}) $$ $$ \vec{F}_{net} = q [ (\vec{v}_A – \vec{v}_B) \times \vec{B} ] $$The relative velocity difference is purely due to rotation:
$$ \vec{v}_A – \vec{v}_B = \vec{\omega} \times (\vec{r}_{A} – \vec{r}_{B}) = \vec{\omega} \times \vec{l} $$Given $\vec{l} = l(\cos\theta \hat{i} + \sin\theta \hat{j})$ and $\vec{\omega} = \omega \hat{k}$:
$$ \vec{v}_A – \vec{v}_B = (\omega \hat{k}) \times l(\cos\theta \hat{i} + \sin\theta \hat{j}) = \omega l (\cos\theta \hat{j} – \sin\theta \hat{i}) $$Substituting this into the force equation with $\vec{B} = B\hat{k}$:
$$ \vec{F}_{net} = q [ (\omega l \cos\theta \hat{j} – \omega l \sin\theta \hat{i}) \times B\hat{k} ] $$ $$ \vec{F}_{net} = q \omega l B [ \cos\theta (\hat{j} \times \hat{k}) – \sin\theta (\hat{i} \times \hat{k}) ] $$ $$ \vec{F}_{net} = q \omega l B [ \cos\theta (\hat{i}) – \sin\theta (-\hat{j}) ] $$2. Calculation of Net Torque
We calculate torque about the center of the dumbbell. The total torque is:
$$ \vec{\tau} = \vec{r}_A \times \vec{F}_A + \vec{r}_B \times \vec{F}_B $$Since $\vec{r}_B = -\vec{r}_A$ and the charge at B is $-q$ (making the force logic symmetric regarding translation), we can simplify by separating translational and rotational contributions.
The torque due to the rotational component of velocity cancels out because the forces form a couple acting outward/inward along the rod axis. The net torque arises from the translational velocity component coupled with the position vectors.
Force due to translation on A: $\vec{F}_{trans, A} = q(\vec{v}_c \times \vec{B})$
Force due to translation on B: $\vec{F}_{trans, B} = -q(\vec{v}_c \times \vec{B})$
These two forces form a couple. The torque is:
$$ \vec{\tau} = (\vec{r}_A – \vec{r}_B) \times \vec{F}_{trans, A} = \vec{l} \times q(\vec{v}_c \times \vec{B}) $$Calculating the cross product:
$$ \vec{v}_c \times \vec{B} = (v_x \hat{i} + v_y \hat{j}) \times B\hat{k} = B(v_y \hat{i} – v_x \hat{j}) $$ $$ \vec{l} = l(\cos\theta \hat{i} + \sin\theta \hat{j}) $$Now, $\vec{\tau} = q \vec{l} \times (\vec{v}_c \times \vec{B})$:
$$ \vec{\tau} = q [ l(\cos\theta \hat{i} + \sin\theta \hat{j}) ] \times [ B(v_y \hat{i} – v_x \hat{j}) ] $$ $$ \vec{\tau} = qlB [ \cos\theta(-v_x)(\hat{k}) + \sin\theta(v_y)(-\hat{k}) ] $$