# Homework Help: Particle track detector to measure the charge of a particle

1. Dec 2, 2017

### Kara386

1. The problem statement, all variables and given/known data
The detector is made of two co-planar layers of silicon two metres apart, and there is a plane of wire chambers halfway between them. The wire chamber can measure a track with precision $100\mu$m. What is the highest momentum a particle can have for which the detector will give the correct sign of the charge with 90% confidence?

There is a 1T field perpendicular to all the planes of the detector, ignore the effects of resolution and assume the track of the particle can be considered orthogonal to both the planes of the detector and the magnetic field.

2. Relevant equations

3. The attempt at a solution
This is going to involve the sagitta in some way. I have a couple of diagrams showing what I think is going on. In the diagrams $s$ is the sagitta and $L$ is the length of the charged particle track. Then I think the statement about precision means the error in the measurement of the sagitta is $\pm 50\mu$m. $B=1$T, $L=2$m. I have also shown that $s=\frac{L^2}{8r}$ and given that $p = eBr$ where $p$ is momentum perpendicular to the field then
$p=\frac{eBL^2}{8s}$
My thoughts on what I need to do are that it might just be some sort of error propagation calculation, or that I need to find the point at which the sagitta is so small that to within error the track might be measured to curve the opposite way to what it really does, which would give the wrong charge of the particle. But I don't know how to actually do either of those things, I am very stuck!

I really appreciate any help. :)

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2. Dec 2, 2017

### Staff: Mentor

It is unclear what "precision of 100 micrometer" means, but I would use it as standard deviation, not as twice the standard deviation.
An actual detector will have a position-dependent uncertainty so you have to integrate over positions, but that is beyond the scope of this problem.

The sign of the charge is the sign of the sagitta. How large does it have to be for a 10% chance to be measured with the wrong sign?

3. Dec 6, 2017

### Kara386

Ah, so I can assume it's distributed normally and use the tables to find the mean such that when the sagitta is zero only ten percent of values are above it. Thank you!