# Homework Help: Electron in a magnetic field problem

1. Dec 2, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I was completely clueless about the hint so I tried the following:

The force acting on the electron is $\vec{F}=e(\vec{v_e}\times \vec{B})$. The speed of electron in the magnetic always stays constant, only the direction changes.

Since $\vec{F}=m_e\vec{\ddot{s_e}}=e(\vec{v_e}\times \vec{B})$. For proton $m_p\vec{\ddot{s_p}}=e(\vec{v_p}\times \vec{B})$

As per the question, the proton follows the same trajectory, hence, $\vec{\ddot{s_e}}=\vec{\ddot{s_p}}=\vec{\ddot{s}}$. This implies that $v_e/m_e=v_p/m_p$. But this gives the wrong answer. Where is the mistake in this approach?

Okay, getting back to the hint, I see that it says to compute the derivative of unit vector along the velocity with respect to s but I am not sure how to use this.

Also, $\vec{ds}=\vec{v_e(t)}dt=|\vec{v_e}|\hat{v_e(t)}dt$ but still I do not see how to proceed.

Any help is appreciated. Thanks!

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2. Dec 2, 2013

### voko

The equality of accelerations is not sufficient. For example, at the sea level, all free falling bodies experience equal accelerations. Yet you can have very different motions depending on the initial velocity. You get identical trajectories only when the initial velocities are identical. But if the velocities in your case are identical, then the proportion you got cannot hold. Contradiction.

But the thing is, accelerations do not have to be equal in identical trajectories. Trajectories are not motions, different motions can still have identical trajectories. For example, you can walk across your room slowly or swiftly: different motions, identical trajectories.

3. Dec 2, 2013

### Saitama

Hi voko! :)

Thanks for the explanation, I understand the mistake now but I am still clueless about approaching the problem. Can I please have a few hints?

4. Dec 2, 2013

### voko

The hint is that you should convert the equation of motion, which is time-dependent, into an equation of trajectory, which is time-independent. The problem already supplies a very strong hint, which is that you use natural parametrization of the curve. All I can add is "use the chain rule".

5. Dec 2, 2013

### Saitama

I honestly cannot proceed.

Do you ask the following, pardon if this looks foolish.

$$\frac{d}{ds}\frac{\vec{v}}{v} =\frac{d\hat{v}}{dt}\frac{1}{ds/dt}=\frac{d\hat{v}}{dt}\frac{1}{\vec{v}}$$

6. Dec 2, 2013

### voko

Start from something simpler. What is $\frac {ds} {dt}$? What is $\vec v = \frac {d \vec r} {dt}$ in terms of s? It is also useful to keep in mind your observation that $v$ is a constant in this problem.

7. Dec 2, 2013

### Saitama

$\frac{ds}{dt}=v$, correct? What is r?

8. Dec 2, 2013

### voko

$\vec r$ is the position of the particle, standard notation if you ask me :)

9. Dec 2, 2013

### Saitama

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \Rightarrow \vec{v}=(dx/dt)\hat{i}+(dy/dt)\hat{j}+(dz/dt)\hat{k}$

But I don't know how I can express $\vec{s}$ in terms of x,y and z. :(

10. Dec 2, 2013

### voko

There is no $\vec s$. $s$ is a scalar defined as shown in #1. It is the natural parameter of the curve (trajectory). You need to convert the diff. eq. of motion (time-dependent) to the diff. eq. of the trajectory (s-dependent). Then find out the condition for equal trajectories.

11. Dec 2, 2013

### Saitama

Since I am still clueless, I am referring a book. Here is the paragraph I am looking at:

Do I have to somehow make use of what is shown in the image?

#### Attached Files:

• ###### para.png
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12. Dec 2, 2013

### voko

$s$ here is a kind of $l$, except it has a special property. Here is another little hint for you: $$\vec v = \frac {d\vec r} {dt} = \frac {d \vec r} {ds} \frac {ds} {dt} = ?$$

13. Dec 2, 2013

### Saitama

I still don't see what $d\vec{r}/ds$ represents, I have never seen anything like this.

14. Dec 2, 2013

### voko

$s$ is the parameter of the curve. The equation of the curve (trajectory) is given by $$\vec r = \vec r(s)$$ The motion along the curve is specified by fixing $s = s(t)$. So $$\vec r = \vec r(s(t))$$ Then you just apply the chain rule.

15. Dec 2, 2013

### Saitama

:uhh:

Don't we end up with the same thing if the quoted equation is differentiated wrt time?

$$\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{ds}\frac{ds}{dt}$$

What do I have to substitute for $d\vec{r}/ds$?

16. Dec 2, 2013

### voko

You keep $d\vec r / ds$. You need the terms independent of $t$ because you want the equation of the curve.

17. Dec 2, 2013

### Saitama

And how do I find the equation of curve?

You stated that the equation of curve is $\vec{r}=\vec{r}(s)$ but I don't see how to figure out the RHS. I am completely blank on this. :(

We have
$$\vec{v}=\frac{d\vec{r}}{ds}v$$

The hint asks to compute $d/ds(\vec{v}/v)$

$$\Rightarrow \frac{d}{ds}\frac{\vec{v}}{v}=\frac{d}{ds}\frac{d\vec{r}}{ds}=\frac{d^2\vec{r}}{ds^2}$$

I will leave for now and go to sleep. Thank you very much voko for the help you have provided so far. :)

18. Dec 2, 2013

### voko

You start with the equation of motion. Which is $$m \frac {d\vec v} {dt} = \vec F$$ Try to express as much as you can in terms of $s$ rather than $t$.

19. Dec 3, 2013

### Saitama

$$\frac{d\vec{v}}{dt}=v\frac{d}{dt} \frac {d\vec{r}}{ds}=v\frac{d}{ds}\frac{d\vec{r}}{ds}\frac{ds}{dt}=v^2\frac{d^2\vec{r}}{ds^2}$$

Is this what you ask me? What should I do next?

20. Dec 3, 2013

### voko

21. Dec 3, 2013

### Saitama

$$\vec{F}=mv\frac{d\vec{v}}{ds}$$

22. Dec 3, 2013

### voko

In #1, you had $\vec F = e \left( \vec v \times \vec B \right)$.

23. Dec 3, 2013

### Saitama

Sorry.

$$\Rightarrow \frac{d^2\vec r}{ds^2}=\frac{e}{mv^2}(\vec{v}\times \vec{B})=\frac{e}{mv}(\hat{v}\times B)$$

So for the same trajectory, we need to have the above expression same for both the proton and electron, right?

24. Dec 3, 2013

### voko

What is $\hat{v}$?

25. Dec 3, 2013

### Saitama

The speed in the magnetic field is constant so I wrote $\vec v$ as $|\vec v|\hat{v}$ where $\hat v$ is the unit vector along the direction of the velocity vector.