# Charge moving through electric and magnetic field

1. Nov 14, 2007

### Supernats

[SOLVED] Charge moving through electric and magnetic field

1. The problem statement, all variables and given/known data
I'm pretty sure my answer for part (a) was correct, but I'll include it just for kicks. Also, I'm sorry if it's wordy, this is my first post and not exactly sure what protocol is, so I'm just copying from the book.

An electron moves in a circular path with radius r=4.00 cm in the space between two concentric cylinders. The inner cylinder is a positively charged wire with radius a = 1.00 mm and the outer cylinder is a negatively charge hollow cylinder with radius b = 5.00 cm. The potential difference between the inner and outer cylinders is $$V_a_b$$ = 120 V, with the wire being at the higher potential. The electric field in the region between the cylinders is radially outward and has magnitude $$E = \frac{V_a_b}{r \ln(\frac{b}{a})}$$.
a) Determine the speed the electron needs to maintain its circular orbit. You can ignore bother the gravitational and magnetic fields of the earth.
b) Now include the effect of the earth's magnetic field. If the axis of symmetry of the cylinders is positioned parallel to the magnetic field of the earth, at what speed must the electron move to maintain the same circular orbit? Assume that the magnetic field of the earth has magnitude 1.30 X 10^-4 T and that its direction is out of the plane of the page in the figure.

2. Relevant equations
For part (a), I used $$\sum F = m a$$, giving me $$q E = \frac{m v^2}{r}$$, which becomes $$v = \sqrt{\frac{r q E}{m}}$$. This gave me 5.39 x 10^12 m/s.

For part (b).
$$F_b = q v \times B$$.
Obviously Newton's second law will come into play again, so $$\sum F = F_B + F_E = q(E + v \times B = m a = \frac{m v^2}{r}$$.

3. The attempt at a solution
I'm basically stumped here. I tried $$\frac{m v^2}{r} = q v B + q E$$ with the quadratic formula but I'm getting a negative discriminant. Dimensional analysis works out, but my numbers don't. (I just put e in for q since we're talking about an electron.)
I got $$v = \frac{r(-e B \pm \sqrt{B^2 e^2 - 4 \frac{m e E}{r}}}{2m}$$.

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Last edited: Nov 15, 2007
2. Nov 15, 2007

### rl.bhat

Earth's magnetic field cannot 1.3x10^4 T. Probably it may be 1.3x10^-4. Taking this value try it again.

3. Nov 15, 2007

### Supernats

Sorry, that was a typo. I calculated using 1.3 G originally and got a negative discriminant. Actually, greater magnetic field should only make it more negative if my equation is correct.

4. Nov 15, 2007

### rl.bhat

While deciding the electron's direction in the electric and magnetic field you have taken into account its sign. Now the equation is mv^2 - reBv- rqE = 0 In this put the magnitude of charge of the electron.Now the discriminant becomes positive.

5. Nov 17, 2007

### Supernats

Of course. I knew I did something stupid like that. Thanks a bunch.