Particle in a circular path due to magnetic field

fight_club_alum

Homework Statement
What will be the radius of curvature of the path of a 3.0-keV proton in a perpendicular
magnetic field of magnitude 0.80 T?
a . 9.9 mm <-- answer
b. 1.1 cm
c. 1.3 cm
d. 1.4 cm
e. 7.6 mm
Homework Equations
v = sqrt( (2 * q * potential) / mass)
mv/bq = r
v = sqrt( (2 * charge of proton * 3000/e) / (mass of proton))
v = 1.893986024 x 10^15
r = ( (mass of proton) * (velocity) ) / ((magnetic field) * (charge of proton))
r = 24715769.68 m

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Orodruin

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• fight_club_alum

ehild

Homework Helper
Relevant Equations: v = sqrt( (2 * q * potential) / mass)
mv/bq = r

v = sqrt( (2 * charge of proton * 3000/e) / (mass of proton))
v = 1.893986024 x 10^15
r = ( (mass of proton) * (velocity) ) / ((magnetic field) * (charge of proton))
r = 24715769.68 m

What is 'e' in the formula for the speed? What is the unit of speed you got?

• fight_club_alum

fight_club_alum

What is 'e' in the formula for the speed? What is the unit of speed you got?
e is the charge of the electron on the constant list in the calculator, which is the same as the proton without the sign

ehild

Homework Helper
e is the charge of the electron on the constant list in the calculator, which is the same as the proton without the sign
Why did you divide by it?

• fight_club_alum

fight_club_alum

The answer comes out if I convert the ev to joule and work normally using the kinetic energy equation 1/2 m v^2 = E
But what is the mistake when I use the other equation?

ehild

Homework Helper
to convert from ev to v. Am I correct?
No. The conversion happened when you multiplied the potential difference with q.
1 eV = Potential difference multiplied by the elementary charge e (which is equal to q in your case)

• fight_club_alum

fight_club_alum

I'm sorry but I don't understand
is v = sqrt( (2q(potential)/ mass)
If yes
isn't ev/electron charge = volt
Thank you

ehild

Homework Helper
The answer comes out if I convert the ev to joule and work normally using the kinetic energy equation 1/2 m v^2 = E
But what is the mistake when I use the other equation?
That equation is wrong as you multiplied and divided the eV-s by the elementary charge.

• fight_club_alum

fight_club_alum

So, the equation should be v = sqrt( (2 * q * ev ) / mass)

ehild

Homework Helper
I'm sorry but I don't understand
is v = sqrt( (2q(potential)/ mass)
If yes
isn't ev/electron charge = volt
Thank you
The equation 'v = sqrt( (2q(potential)/ mass)' is wrong. Instead of 'potential' it must be energy, but you have the energy in eV-s here, which converts to joules, if you multiply the potential value (3000) with the electronic charge.
If an electron or proton travels across a potential difference of U Volts, it gains e U kinetic energy. But we say that it gained U eV energy. So we measure energy by the potential difference which would accelerate the electron/proton to that energy.

• fight_club_alum

fight_club_alum

The equation 'v = sqrt( (2q(potential)/ mass)' is wrong. Instead of 'potential' it must be energy, but you have the energy in eV-s here, which converts to joules, if you multiply the potential value (3000) with the electronic charge.
If an electron or proton travels across a potential difference of U Volts, it gains e U kinetic energy. But we say that it gained U eV energy.
thank you
so the right one should be
v = sqrt ( (2 * e * (3000e) / (Mass of proton) ) <--- e is just the charge of the proton

Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
The equation 'v = sqrt( (2q(potential)/ mass)' is wrong.
It is not wrong, but eV is not a unit of electric potential, it is an energy unit, ie, the value given is the charge multiplied by the accelerating potential. In fact, there is no reason to even talk about an accelerating potential here as it does not matter how the proton was accelerated. The only thing that matters is that it has a given kinetic energy.

• fight_club_alum

ehild

Homework Helper
thank you
so the right one should be
v = sqrt ( (2 * e * (3000e) / (Mass of proton) ) <--- e is just the charge of the proton
No. $energy=0.5 mv^2 = 3000e$
So $v=\sqrt{2*3000e/m}$

• fight_club_alum

fight_club_alum

When I do this
v = sqrt ( 2 * e * 3000/ mass of the proton) <---- where e is the charge of the proton and 3000 is the ev I get the right answer
So normally I should use (ev) not v

fight_club_alum

great! thank you so much
Sorry if I confused the whole forum

ehild

Homework Helper
When I do this
v = sqrt ( 2 * e * 3000/ mass of the proton) <---- where e is the charge of the proton and 3000 is the ev I get the right answer
So normally I should use (ev) not v
Yes, you get the speed from the kinetic energy, and you can measure energy in eV units, but it should be converted to joules, by multiplying the electron-volts by e. Here e is the elementary charge, not the charge of the particle in question.
If it was an oxygen ion with charge 2e, and energy 3000 eV, the energy in joules would be 3000*e again.

• fight_club_alum

fight_club_alum

Yes, you get the speed from the kinetic energy, and you can measure energy in eV units, but it should be converted to joules, by multiplying the electron-volts by e.
Thank you
but
if I did this to convert ev to j
sqrt(2 * e * (3000e) / mass) <---- where e is the charge of a proton and 3000e is the ev to joule conversion
I get the wrong answer

ehild

Homework Helper
Thank you
but
if I did this to convert ev to j
sqrt(2 * e * (3000e) / mass) <---- where e is the charge of a proton and 3000e is the ev to joule conversion
I get the wrong answer
You multiplied by e twice. v=sqrt(2*energy/mass) and energy (in joules) = e*energy (in electron-volts)

• fight_club_alum

fight_club_alum

Oh, I see what I've been doing
Thank you and sorry for taking too long

ehild

Homework Helper
You are welcome • fight_club_alum

"Particle in a circular path due to magnetic field"

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