0
$\begingroup$

Sorry for asking an exercise question here, but I'm really having difficulty figuring this one out.

Quick summary of the problem: a charged particle moving in a uniform magnetic field will experience a constant horizontal "drift" in a direction parallel to the magnetic field, while having circular motions with an axis parallel to the magnetic field. Is it possible to impose an external, constant force field (say, an electric field) such that this "drift" is cancelled out?

The problem, taken from Professor David Tong's Lectures on Dynamics and Relativity:

A particle of mass $m$, charge $q$ and position $\vec{x}$ moves in a constant, uniform magnetic field $\vec{B}$ which points in a horizontal direction. The particle is also under the influence of gravity, $\vec{g}$, acting vertically downwards. Write down the equation of motion and show that it is invariant under translations $\vec{x}\to\vec{x}+\vec{x_{0}}$. Obtain $$\dot{\vec{x}}=\alpha\vec{x}\times\hat{n}+\vec{g}t+\vec{a}$$ where $\alpha=\frac{qB}{m}$, $\hat{n}$ is a unit vector in the direction of $\vec{B}$ and $\vec{a}$ is a constant vector. Show that, with a suitable choice of origin, $\vec{a}$ can be written in the form $\vec{a}=a\hat{n}$. By choosing suitable axes, show that the particle undergoes a helical motion with a constant horizontal drift. Suppose that you now wish to eliminate the drift by imposing a uniform electric field $\vec{E}$. Determine the direction and magnitude of $\vec{E}$.

My solution so far:

Equation of motion: $$ m\ddot{\vec{x}}=q\dot{\vec{x}}\times\vec{B}+m\vec{g} \tag{1} $$ i) Define a coordinate system in which $$\vec{x}'=\vec{x}+\vec{x}_{0}\tag{2}$$
ii) Then $\dot{\vec{x}'}=\dot{\vec{x}}$ and $\ddot{\vec{x}'}=\ddot{\vec{x}}$ so the equation of motion is invariant.

iii) Integrating our equation of motion with regard to time will yield $$\dot{\vec{x}}=\alpha\vec{x}\times\hat{n}+\vec{g}t+\vec{a}\tag{3}$$

iv) Showing that $\vec{a}$ can be expressed in the form $a\hat{n}$:
Let us substitute this equation with $$\vec{x}=\vec{x}'-\vec{x}_{0}\tag{4}$$ This yields $$\dot{\vec{x}'}=\alpha\vec{x}'\times\hat{n}+\vec{g}t+(\vec{a}-\alpha\vec{x}_{0}\times\hat{n})\tag{5}$$ But we also know the following because of the invariance of the equation of motion upon a linear translation: $$\dot{\vec{x}'}=\alpha\vec{x}'\times\hat{n}+\vec{g}t+\vec{a}'\tag{6}$$
where $\vec{a}'$ is a constant and can be expressed as having a component perpendicular to $\hat{n}$ and one parallel to $\hat{n}$. For some values of $\vec{x}_{0}$, $\alpha\vec{x}_{0}\times\hat{n}$ equals the component of $\vec{a}'$ that is perpendicular to $\hat{n}$ so $\vec{a}$ is parallel to $\hat{n}$ and can be expressed as $\vec{a}=a\hat{n}$.

WLOG, $\vec{B}=B\hat{i}$, $\vec{g}=-g\hat{k}$. Returning to the equation describing $\dot{\vec{x}}$ and expanding the cross product yields $$\dot{\vec{x}}=\alpha(z\hat{j}-y\hat{k})-gt\hat{k}+a\hat{i}\tag{7}$$
The first two terms describe circular motion, the third term makes it an accelerating helical motion, and the fourth term provides a constant horizontal drift (in the direction of the B-field).

To find the E-field, we insert it into our original equation of motion and follow the same steps as above to get $$\dot{\vec{x}}=\alpha\vec{x}\times\hat{n}+\vec{g}t+\vec{a}+q\vec{E}t\tag{8}$$
I'm guessing the resulting motion will have a rotated axis of 'rotation' from the previous motion, and also that the solution hinges on solving for $\vec{a}$. Also, I've observed that at this point, instead of considering gravitational and electric fields separately, they can be computed as a general force field (since they can add). And from here I'm stuck.

My line of reasoning is that the force field can be broken up into a component perpendicular to the magnetic field and a component parallel to the magnetic field. The parallel component wouldn't help eliminate the drift because it would just accelerate the particle away. But the perpendicular component is demonstrated by the gravitational field in the problem already, where the horizontal drift component still exists. I must have overlooked something conceptually here.

Any help or hints would be appreciated. Thanks!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.