# Linear accelerator physics problem

1. Apr 24, 2004

### carnom

Ive been working on this problem for a few hours and cant get the last few parts.

In a linear accelerator, protons are accelerated from rest through a potential difference to a speed of approximately 3.1 X 10^6 meters per second. The resulting proton beam produces a current of 2 X 10^-6 ampere.

I found part B and D (with respect to the answer of part A), but i cant get part A, which is: Determine the potential difference through which the protons were accelerated.

I also cant get part C, which is: The proton beam enters a region of uniform magnetic field B that causes the beam to follow a cemicircular path. Determine the magnitude of the field that is required to cause an arc of radius .1 meter.

Any help is appreciated. Thanks!

Last edited: Apr 24, 2004
2. Apr 24, 2004

### Blistering Peanut

For part C:

centripical force = (mv^2)/r
this equals force on a current carrying conductor in a magnetic field = BIl

BIl = mv^2/r

for an arc l = r .theta where theta is the angle in radians. For a semi circle theta = pi radians

so the magnetic flux density of that field should equal

$$B = \frac {m v^2}{ I r^2 \pi}$$

3. Apr 24, 2004

### Blistering Peanut

oh yeah for A

from the definition of potential difference
W=QV
for an electron/proton qV = 1/2mv^2
so V = 1/2mv^2/q

4. Apr 24, 2004

### carnom

Thanks, but for A you put a q in the formula, but the charge is not given and i have no idea how to find it. A bit more help would be appreciated. Thanks again. Also, m= 1.67 X 10^-27, right?

Last edited: Apr 24, 2004
5. Apr 24, 2004

### Blistering Peanut

I don't know the mass of the proton without looking it up, but that sounds about right. Did you get the correct answer for part C?

The charge on the proton is the same as that on the electron, 1.6 * 10^-19 C

Last edited: Apr 24, 2004