Finding mass of ions in magnetic field

[s-f]

A device is used to measure the charge-to-mass ratio (e/m) of ions by accelerating them using an electric field described by a potential difference , and once they have acquired kinetic energy of known amount, they enter a uniform magnetic field. The particles move in a plane perpendicular to the magnetic field. What is the mass of the ion that enters the detector?

Strength of magnetic field = 0.8 T
d = 6.00 cm
ΔV = 100 V
Ion charge = 1.6 x 10^-19 C

I'm not really sure how to set up an equation here. I tried using qΔV = 1/2 mv² and F = ma = qvB sin theta to solve it but I couldn't figure out how to solve for m here. Is there another formula I should use?

 

[voko]

It seems some information is missing. What is d? A sketch of the setup would help, too.

 

[truesearch]

From your first equation you can write v^2 = qV/m.
In the magnetic field the path is circular so Bqv = mv^2/r
You can combine these equations to give m = B^2xqxr^2/2V
Can you see how to do that?
If d in your question is the DIAMETER of the path in the magnetic field you should be able to solve it.
Have a go and come back if you get stuck.

 

[s-f]

Sorry this is the image https://mail-attachment.googleusercontent.com/attachment/u/0/?ui=2&ik=d3e9366f37&view=att&th=138f2cfebf63fad8&attid=0.1&disp=inline&realattid=1409394640159768576-1&safe=1&zw&saduie=AG9B_P-_1Tbsc3zXwuOtwk13kB5g&sadet=1344103689147&sads=0rBGq4MW_H8NzAdDPKWjvF6cP-g

@truesearch - No I didn't get that equation at all when I tried to do it. Can you explain how you came up with that? I know it's correct because I plugged in the variables and it came out to 4.6 x 10^-25 kg which is the correct answer.

 

[voko]

The picture is inaccessible.

 

[s-f]

Hopefully you can click to make it bigger. Truesearch is right that d is the diameter and that the magnetic field in the path is circular.

 

[voko]

What is the force acting on a charge in magnetic fields? What is the acceleration in uniform circular motion?

 

[s-f]

Here is what I tried to do originally:

F = qvB = ma
Since a = v²r, qvB = m(v²r)

qΔV = 1/2 mv², so v = sqrt (2ΔV)/m. I tried to plug this in as v for the previous equation but I must have messed up algebraically because I did not get the final equation Truesearch posted earlier.

 

[voko]

Since a = v²r

This is not the correct formula for acceleration in uniform circular motion!

 

[s-f]

This is not the correct formula for acceleration in uniform circular motion!

You're right; I typed it wrong. I was using a = v²/r though which I think it correct?

 

[voko]

From kinetic energy, you get $v^2 = \frac {2 q V} {m}$. Plug that into the $F = ma$ formula and you should get what you need.

 

[s-f]

Hmm I still think I'm fudging the algebra (terrible at math). I ended up getting:

m = qBr/sqrt (2qV/m)

 

[voko]

Get rid of the root by squaring.

 

[s-f]

Still not getting m = B^2xqxr^2/2V or 4.6 x 10^-25. I ended up not getting q in my numerator which makes no sense.

 

[voko]

You wrote $$m = \frac {q B r} {\sqrt {\frac {2 q V} {m}}}$$

Squared, that becomes $$m^2 = \frac {q^2 B^2 r^2} {\frac {2 q V} {m}} = m \frac {q B^2 r^2} {2 V}$$

You really should have more algebraic practice, these are very simple transformations, things can get much worse.

 

[s-f]

Ok I see, I was making a dumb mistake in the last step - thanks for explain that part.

And yeah I know I'm awful at math but this is the only physics class I need to take to finish high school and it's over in a week :) Not planning on doing physics in college!

 

[truesearch]

Well done s-f
This is quite tricky algebra