# Charged Particle in Magnetic field calculation

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1. Dec 26, 2016

### Matthew Clisby

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

Basically having a problem when trying to theoretically calculate the mass of alpha particle in a charged field.

r is the unknown radius of curvature
m = mass of an alpha particle (6.646 *10^-27 kg)
v = velocity of an alpha particle immediatly after decay (1.381 *10^7 m/s)
q = charge of an alpha particle (3.2 *10^-19 C)
B = magnetic field strength (0.005 Tesla)

2. Relevant equations

r=mv/qB

3. The attempt at a solution

Answer should be around 1cm but instead value is around 79cm. What's going wrong, am I forgetting something obvious. 1cm is published everywhere and was similar to my experimental results.

Thanks

Last edited: Dec 26, 2016
2. Dec 26, 2016

### Stavros Kiri

What is r? Circular orbit? State more accuratelly all data and unknown ...

3. Dec 26, 2016

### Stavros Kiri

+ check the equation and in the proper unit system ...

4. Dec 26, 2016

### Matthew Clisby

Thanks Stavros I've updated it would and it checks out fully with the SI unit of equations, is there any value I've used that seems blindingly wrong to you?

5. Dec 26, 2016

### Stavros Kiri

Seems ok now. Values also seem ok and the equation is correct (comes from Lorentz force and circular motion). Did you convert cm to m properly (for SI units - not CGS)? Perhaps 79cm is 0.79 or something ... (I didn't do the calculation). You check it first, all units converted properly in SI ...

6. Dec 26, 2016

### Matthew Clisby

Yeah, annoying thing is that I've done exactly that but the answer just comes out wrong..., even then 79cm is quite a way off of 1cm

7. Dec 26, 2016

### Staff: Mentor

What is the source of your data for the velocity of the alpha particle and the strength of the magnetic field? Those two would appear to be the only things in your scenario that are not intrinsic properties of the particle.

8. Dec 26, 2016

### Matthew Clisby

The field is just an exemplar of a lowly induced field and the velocity calculation is considerably more complex: (see picture)

This value is also similar to one found on Wikipedia (I know the source is awful I couldn't find it anywhere else)

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• ###### Screen Shot 2016-12-26 at 21.49.41.png
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9. Dec 26, 2016

### Staff: Mentor

The depicted velocity calculation is a bit overly detailed for the present purpose, but gives a value that seems reasonable to me. It essentially apportions the energy released according to the masses of the daughter particles (radium nucleus and alpha particle). That just leaves the magnetic field to wonder about.

For the velocity attributed to the alpha particle, what magnetic field strength would result in a trajectory radius of about 1 cm? It would have to be much larger than 0.005 T.

Can you point us to an example of this published value?

10. Dec 26, 2016

### Stavros Kiri

I agree. Your data can be the only source of the inconsistency.