Proton vs electron, find the magnetic field for the electron

[Alice7979]

Homework Statement

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.30 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

Homework Equations

q electron= -1.60E-19 C
q proton= 1.60E-19 C

The Attempt at a Solution

If the velocity and radius are the same they cancel out but I can't set them equal to each other because the Bs are different, right?
F1= q1v1B1
F2= q2v2B2
What I'm i forgetting?

 

[tnich]

Homework Statement

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.30 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

Homework Equations

q electron= -1.60E-19 C
q proton= 1.60E-19 C

The Attempt at a Solution

If the velocity and radius are the same they cancel out but I can't set them equal to each other because the Bs are different, right?
F1= q1v1B1
F2= q2v2B2
What I'm i forgetting?

F=ma

 

[Alice7979]

ELECTRON: 9.11E-31(v^2/r) = (-1.60E-19)(B)(v)
PROTON: 1.673E-27(v^2/r) = (1.60E-19)(.3)(v)
9.11E-31(v^2/r) / (-1.60E-19)(B)(v) = 1.673E-27(v^2/r) / (1.60E-19)(.3)(v)
Belectron: -1.6E-4

 

[Chandra Prayaga]

Your equations for the force are for the magnitudes only. There should be no negative signs for the electron charge in it.

 

[ZapperZ]

@Alice7979 : I have no idea what you mean by "... they cancel out.. ". It doesn't appear that you know how to tackle even the first part of the problem, and your follow-up post doesn't help at all.

1. Write down the Lorentz force equation (in symbolic form). Then at least we know that you know where to start.

2. Using the Lorentz force equation, find the magnetic force acting on the proton first.

3. Use this magnetic force as the centripetal force, and find the radius of the circular motion that the proton makes.

Now, can you prove that you are able to do the above first? Ignore the second part for the moment, because if you can't do this straightforward part, you have a bigger issue in understanding the physics here than just solving this problem.

Zz.

 

[tnich]

@Alice7979Then at least we know that you know where to start.

Your comments seem to ignore that Alice7979 has already solved the problem in post #3.

 

[ZapperZ]

Your comments seem to ignore that Alice7979 has already solved the problem in post #3.

But I don't know if this is simply a plug-and-chug or if this person actually understood what needs to be done. That was why I stated that it is unclear, at least to me, if he/she is aware of what is going on.

If this is solved and the OP knows exactly what is involved, then kumbaya, we are all good!

Zz.

 

[Alice7979]

But I don't know if this is simply a plug-and-chug or if this person actually understood what needs to be done. That was why I stated that it is unclear, at least to me, if he/she is aware of what is going on.

If this is solved and the OP knows exactly what is involved, then kumbaya, we are all good!

Zz.

Yea, I wasn't sure if my work was right

F=qvB ----> F= (1.60E-19)v(.30)

I don't know if I should try to find v and I don't remember how but if qB=mv/r, can I solve for v/r and do it that way?

 

[Charles Link]

Yea, I wasn't sure if my work was right

F=qvB ----> F= (1.60E-19)v(.30)

I don't know if I should try to find v and I don't remember how but if qB=mv/r, can I solve for v/r and do it that way?

Yes, you can write the expression for $\frac{v}{r}=\frac{qB}{m}$. Since $\frac{v}{r}$ is the same for both cases, you can set them equal. Since you already solved it, I'm just showing what is a simpler way: $\frac{v}{r}=\frac{q_1 \, B_1}{m_1}=\frac{q_2 \, B_2}{m_2}$. The q's (absolute value) are the same,(we're really just working with absolute values throughout this problem), so they cancel for the expression $\frac{q_1 \, B_1}{m_1}=\frac{q_2 \, B_2}{m_2}$. $\\$ (If the proton orbits clockwise, the electron will orbit counterclockwise if the sign of $B$ is kept the same). $\\$ Solve for $B_2$. You don't need to actually solve for the numerical value of $\frac{v}{r}$, but you could if you want.

 

[Alice7979]

(9.11E-31)(1.60E-19)(.3)/(1.60E-19)(1.673e-27) =1.6E-4
I got it right but the homework program doesn't work that well so I didn't really get it right had to have 3 sig figs :( Thank you though