Electron drift caused by a magnetic field gradient

1. Dec 22, 2013

eoghan

1. The problem statement, all variables and given/known data
The magnetic field of the Earth it's approximately $B=3\times10^{-5}T$ at the equator and diminishes with the distance from the center of the Earth as $1/r^3$, as a dipole. Consider a population of electrons on the equatorial plane with energy 30keV, at 5 earth radius from the center of the Earth.
Considering only the drift caused by the gradient of the magnetic field (i.e. ignore the curvature of the field lines) determine the velocity drift.

2. Relevant equations
$$v_d=-\frac{\epsilon_\bot}{B}\frac{\nabla B\times B}{qB^2}$$

3. The attempt at a solution
$$|v_d| = \frac{3\times 10^4eV}{(3\times 10^{-5}T)^2}\frac{1}{(5\cdot6371\times10^3m)^3} \approx 10^{-15} m/s$$

Is this result correct? Because it seems to me to be too low....moreover the next question asks to compute how many time will take for the electron to go around the earth circumference.
Thank you very much for your help

2. Dec 22, 2013

Staff: Mentor

Your units do not match, the result cannot be right. I don't see how you got those values based on the formula.

3. Dec 23, 2013

eoghan

Yes, I made an error, the energy in the numerator is not in eV, but in Joule, because in the formula there is
$$\frac{\epsilon_\bot}{q}=\frac{3\times10^4eV}{e}=3\times10^4J$$
Are the units now correct?
Anyway, the result doesn't change

4. Dec 23, 2013

Staff: Mentor

eV and Joule are both energies, you have to keep the conversion in mind but that is not the issue I meant: the fraction just does not give a velocity.
It's like answering a question for a velocity with "5m". 5m is not a velocity, it cannot be the right answer.

Your new formula has wrong units as well, an energy divided by a charge is not an energy.

5. Dec 24, 2013

eoghan

Ok... I understand where I am wrong, I took as the gradient of $B$ the value $1/r^3$ which is not correct.
To find the gradient I would do like that. The B field has the form $k/r^3$, where $k$ is a constant. To find this constant I know that at the equator $B_{eq}=k/r_{eq}^3=3\times 10^{-5}T$.
So $k=B_{eq}r_{eq}^3$. The gradient is thus $\nabla B=-\frac{B_{eq}r_{eq}^3}{r^4}$.
What do you think? Is this correct?

6. Dec 24, 2013

Staff: Mentor

There is a factor 3 missing (the gradient is the derivative of the field strength), but apart from that it looks good.

7. Dec 29, 2013

eoghan

Thank you very much for your help!