- #1
akhi999
- 1
- 0
- Homework Statement
- In a region of space there is a uniform and constant magnetic field along the z axis ##\vec{B}=B\hat{z}##. Two point-like charges of equal charge ##q## and mass ##m## are fixed at the far ends of a massless rod of length ##2R##. Initially, the rod is at rest in the magnetic field and it is perpendicular to it. At t=0 the charges are given an instantaneous speed along the z axis. The speeds have the same magnitude, equal to ##v##, but they point in opposite direction.
In the subsequent motion, what is the minimum angle between the rod and the magnetic field?
- Relevant Equations
- Biot Savart law, Lorentz force
I have some difficulties in solving this problem. This is what I did.
I wrote down the equation of motion for the masses. For the first point
\begin{equation}
m\ddot{\textbf{r}}_1=\textbf{F}_1=q\dot{\bar{\textbf{r}}}_1\times \textbf{B}+\frac{\mu_0q^2}{4\pi}\frac{1}{\lvert\textbf{r}_1-\textbf{r}_2\rvert^2}\dot{\bar{\textbf{r}}}_1\times(\dot{\bar{\textbf{r}}}_2\times\textbf{r}_1)
\end{equation}
For the second mass we can write a similar equation, with the subscripts 1 and 2 switched.
To derive that, I took into account the fact that when one of the charges moves, it generates a magnetic field (Biot-Savart law). So the other charge feels a Lorentz force due to the initial field ##B## and the new field.
Analyzing the forces, it seems to me that they are equal and opposite. So there is a torque that makes the rod rotate in a circumference, and there is an angular acceleration, namely the angular velocity is not constant.
Hence, I can set
\begin{equation}
\lvert \textbf{r}_1 \rvert=\lvert \textbf{r}_2 \rvert=R, \hspace{3mm}\lvert\textbf{r}_1-\textbf{r}_2\rvert=2R, \hspace{3mm}\dot{\bar{\textbf{r}}}_1=-\dot{\bar{\textbf{r}}}_2=R\dot{\theta}\hat{e}_{\theta}
\end{equation}
Then I used ##\tau=I\ddot{\theta}##, where ##\tau## is the torque, ##I=2mR^2## the moment of inertia around the origin, which I put in the middle of the rod. Since ##\tau=\textbf{r}_1\times \textbf{F}_1+\textbf{r}_2\times\textbf{F}_2##, I have to take the cross product of the right -hand side of the equation of motion (which is the force)with ##\bar{\textbf{r}}_1## to get ##\tau##. If I do this, I get a differential equation for θ, but is it not linear . It is something like
\begin{equation}
\ddot{\theta}=\frac{q}{mR}\dot{\theta}\cos{\theta}
\end{equation}
In principle I should solve now for ##\theta##. But I am not sure if this is correct.
I wrote down the equation of motion for the masses. For the first point
\begin{equation}
m\ddot{\textbf{r}}_1=\textbf{F}_1=q\dot{\bar{\textbf{r}}}_1\times \textbf{B}+\frac{\mu_0q^2}{4\pi}\frac{1}{\lvert\textbf{r}_1-\textbf{r}_2\rvert^2}\dot{\bar{\textbf{r}}}_1\times(\dot{\bar{\textbf{r}}}_2\times\textbf{r}_1)
\end{equation}
For the second mass we can write a similar equation, with the subscripts 1 and 2 switched.
To derive that, I took into account the fact that when one of the charges moves, it generates a magnetic field (Biot-Savart law). So the other charge feels a Lorentz force due to the initial field ##B## and the new field.
Analyzing the forces, it seems to me that they are equal and opposite. So there is a torque that makes the rod rotate in a circumference, and there is an angular acceleration, namely the angular velocity is not constant.
Hence, I can set
\begin{equation}
\lvert \textbf{r}_1 \rvert=\lvert \textbf{r}_2 \rvert=R, \hspace{3mm}\lvert\textbf{r}_1-\textbf{r}_2\rvert=2R, \hspace{3mm}\dot{\bar{\textbf{r}}}_1=-\dot{\bar{\textbf{r}}}_2=R\dot{\theta}\hat{e}_{\theta}
\end{equation}
Then I used ##\tau=I\ddot{\theta}##, where ##\tau## is the torque, ##I=2mR^2## the moment of inertia around the origin, which I put in the middle of the rod. Since ##\tau=\textbf{r}_1\times \textbf{F}_1+\textbf{r}_2\times\textbf{F}_2##, I have to take the cross product of the right -hand side of the equation of motion (which is the force)with ##\bar{\textbf{r}}_1## to get ##\tau##. If I do this, I get a differential equation for θ, but is it not linear . It is something like
\begin{equation}
\ddot{\theta}=\frac{q}{mR}\dot{\theta}\cos{\theta}
\end{equation}
In principle I should solve now for ##\theta##. But I am not sure if this is correct.
Last edited:
