Charges, rod and magnetic field

  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
haruspex
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Never studied the subject to this level, but can you solve it using energy?
The rotating rod generates a magnetic dipole. As the angle of this field changes, there is a change in the stored magnetic energy of the two fields. This depends on both the angular velocity of the rod and its orientation. Meanwhile, the KE depends only on the velocity, and the sum of the two is constant. Maybe there's a way to extract a minimum possible angle from that... but I'm not hopeful.
 
  • #3
rude man
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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.
Some interpretation may be necessary.
I assume the velocity is not implied to remain constant at v as the charges move along the rod. I would assume the velocity is v at the moment it's imparted, and not afterwards. So what is the minimum angle needed to maintain an initial velocity v in the right direction over a very short interval of time?
Given that interpretation, I would forget Biot-Savart. and the mag field generated by the charges. Also forget torques. The charge repulsion force does not impart a torque about the rod's center.
Think Coulomb vs. Lorentz forces on q.
 
  • #4
BvU
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:oldconfused:

the charges move along the rod
Doesn't go well with
Two point-like charges of equal charge ##q## and mass ##m## are fixed at the far ends of a massless rod of length ##2R##.
 
  • #5
rude man
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:oldconfused:

Doesn't go well with
At t=0 they move with velocity v, no longer fixed, although I assumed an infinitesimal displacement from +/- R.

Didn't notice the mass m though. Will reconsider.
 
  • #6
haruspex
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At t=0 they move with velocity v,
As in, the rod initially along the X axis say, starts to rotate in the XZ plane at rate v/R.
 

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