Magnetism Series #2

Assume that the rod is hinged at the origin and that the central rod lies in the first quadrant. Let the coordinates of each of the charged particles be $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Let the distance of each of the charges be $r$ from the central rod along the perpendicular rod. The coordinates of the particles are then:

$x_1 = L\cos(\omega t) + r\sin(\omega t)$ $y_1 = L\sin(\omega t) - r\cos(\omega t)$

$\vec{r}_1 = x_1 \ \hat{i} + y_1 \ \hat{j}$ $\vec{v}_1 = \frac{d \vec{r}_1}{dt}$ $\vec{a}_1 = \frac{d \vec{v}_1}{dt}$

$x_2 = L\cos(\omega t) - r\sin(\omega t)$ $y_2= L\sin(\omega t) + r\cos(\omega t)$

$\vec{r}_2 = x_2 \ \hat{i} + y_2 \ \hat{j}$ $\vec{v}_2 = \frac{d \vec{r}_2}{dt}$ $\vec{a}_2 = \frac{d \vec{v}_2}{dt}$

The forces acting on each of the particles are the electrostatic force along the perpendicular rod, the magnetic force and the reaction force due to the rod (directed along the central rod. The electrostatic force on one of the charge 1 is:

$\vec{F}_e = \frac{Kq^2}{4r^2} \left(\sin(\omega t) \ \hat{i} - \cos(\omega t) \ \hat{j} \right)$

The magnetic force on charge 1 is:

$\vec{F}_b = q\left(\vec{v}_1 \times (-B \ \hat{k})\right)$

The magnitude of the reaction force on charge 1 is $N_1$ . The reaction force vector is:

$\vec{F}_{r1} = N_1 \left(\cos(\omega t) \ \hat{i} + \sin(\omega t) \ \hat{j} \right)$

Finally, applying Newton's second law for charge 1 gives:

$m \vec{a}_1 = \vec{F}_e+\vec{F}_b+\vec{F}_{r1}$

Plugging in all expressions and separating the $x$ and $y$ components of the force gives us two equations. Using these two equations, one can solve for two unknown variables which are $N_1$ and $\ddot{r}$ . After a lot of simplification, the expression for $\ddot{r}$ comes out to be:

$\ddot{r} = \frac{Kq^2-4Bqr^3\omega+4mr^3\omega^2}{4mr^2}$

Looking at the above expression, the particle achieves equilibrium when $\ddot{r}=0$ . This means:

$Kq^2-4Bqr^3\omega+4mr^3\omega^2=0$ $\implies r^3 = \frac{q^2}{16 \pi \epsilon_o \omega (qB - m\omega)}$ $\implies r = \left(\frac{q^2}{16 \pi \epsilon_o \omega (qB - m\omega)}\right)^{1/3}$

Now to find the angle $\theta$ , simple trigonometry can be applied to obtain the expression:

$\tan{\theta} = \frac{r}{L}$ $\implies\boxed{ \tan{\theta} = \left(\frac{q^2}{16 \pi \epsilon_o \omega L^3 (qB - m\omega)}\right)^{1/3}}$

These problems are not difficult but are very tedious to work out. As an exercise for yourself, do the same steps for charge 2. You should get the same expression for $\ddot{r}$ when you do so.