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Radius of the path of a charged particle in a mass spectrometer

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider the mass spectrometer shown schematically in Figure P19.30. The electric field between the plates of the velocity selector is 850 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.910 T. Calculate the radius of the path in the system for a singly charged ion with mass m = 2.18 x 10-26 kg. (Hint: The figure is of a mass spectrometer. Charged particles are first sent through a velocity selector. They then enter a region where a magnetic field B_0 (directed inward) causes positive ions to move in a semicircular path and strike a photographic film at P)


    2. Relevant equations

    r = mv/qB

    3. The attempt at a solution

    I know that I am suppose to use the above formula but I am having a problem finding v. I am not sure what formula I am suppose to use to find the velocity with the given number.

    Any help would be appreciated
  2. jcsd
  3. Mar 15, 2007 #2
    Here, you have to use the Lorentz force equations. [tex]F=q(v\mult B + E)[/tex]. Since the magnetic force here is a variable (velocity varies due to the electric force), you will have to find the instantaneous force in the x and the y directions, which will give you a DE.

    This DE will be of the form [tex]\frac{d^2x}{dt^2}=-\omega ^2x[/tex], which represents SHM. After you do all this, you can finally find the velocity of the ion when it leaves the tube. You use that v in the expression you quoted to find the radius of the circle.
  4. Mar 15, 2007 #3


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    The equation should read ( equation was correct but the tex didn't work )

    [tex]F=q(v\times B + E)[/tex]
  5. Mar 15, 2007 #4
    The formula given is a bit confusing. If I use F = q(vxB + E) I am still stuck because I do not know what F is. Would I have to set two equation together?

    like: qvB = q(vxB + E)?

    I just realize that will not work either since I do not know what x is.

    Also I dont think we ever cover the Lorentz force equation and I don't understand DE.

    These are the formula that i know:

    r = mv/qB

    B = u_oI/2pir

    EB||deltaL = u_0I

    F/L = u_0 I1 I2 / 2pid

    torque = BIAsintheta

    F = qvBsintheta

    F = BILsintheta
  6. Mar 15, 2007 #5


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    BunDa, you misunderstand. That is a vector equation and the 'X' is the vector cross-product.

    [tex]\vec{F}=q(\vec{v}\times \vec{B} + \vec{E})[/tex]

    The equations you've got will do the job as well. Draw a diagram so you you can picture the directions of the charge and the E and B fields.
  7. Mar 15, 2007 #6
    I am still a bit confuse and not really sure how i would draw a diagram for this. Today class did not even cover this either.

    do i use this and solve for v? qvB = q(vB + E)? I know i just removed the x.

    Thanks for the help so far, but when it comes to physics it takes me forever to understand it.
  8. Mar 15, 2007 #7


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    Last edited by a moderator: May 2, 2017
  9. Mar 16, 2007 #8
    No. The expression you're using, qvb is a part of the lorentz force. The total lorentz force is magnetic+electric, where E is the electric field. You would want to use vectors here while doing the maths, cause that would make your calculations easier. Use vectors for expressing the magnetic field and the electric field as well.

    Then, you'll see that the x velocity component is dependend on d(vz)/dt, and similarly the z component is dependent on d(vx)/dt. Youll have to differentiate either of the expressions and substitute the remaining expression for the DE. There is a general solution y=Asinwt. Then youll have to compare equations to find A and w.

    Give it a shot. Lets see the work. We'll help you out.
  10. Mar 16, 2007 #9
    each time i read this i get more confuse.

    I think I will wait till Tuesday to give this another try since I emailed the teacher and she said she will go over this with the class since I asked people in my class and they are all stuck at it also.

    Thanks for the help and explanation.

    I will be back if I still get stuck after the professor explanation.
  11. Mar 16, 2007 #10
    This is much simpler than it seems:

    The particle travels in a straight line with constant velocity while in the magnetic and electric fields. These two forces cancel out. They will only cancel when the particle has a certain velocity because magnetic force is proportional to velocity. You can then solve for radius once you sum up the forces and put the velocity in terms of the electric field
  12. Mar 17, 2007 #11
    No way. How can it? Both the forces are perpendicular to each other!!!
  13. Mar 17, 2007 #12
    Let us assume that at a certain instant of time, the particle has a velocity [tex]V(i,j,k)=v_xi+v_yj+v_zk[/tex].

    Lets choose an axis such that [tex]B=B_0k[/tex] and [tex]E=E_0i[/tex].

    Let the charge on the ion be q.

    Therefore, the net force is given by:
    [tex]F(i,j,k)=q(V(i,j,k)\times B_0k +E_0i[/tex]
    [tex]F(i,j,k)=q((v_xi+v_yj+v_zk)\times B_0k +E_0i[/tex]
    Taking the cross product, we get:
    [tex]F(i,j,k)= q((v_yB_0+E_0)i +(B_0v_x)(-j))[/tex]
    Now, notice here that there is no force acting along the z axis (Along the Magnetic field).

    Therefore, the force along the x axis is given by [tex]F_x=q(v_yB_0+E_0) [/tex]
    And the force along the y axis is given by [tex]F_y=-qB_0v_x[/tex]

    Now, notice here, that the force along the x axis depende on the velocity along the y axis and vice verca.

    Rewriting the above equations as:

    and writing the expression for the force along the y axis in a similar manner to find the acceleration along the respective axis:


    Differentiating the first equation

    Input the value of [tex]\frac{dv_y}{dt}[/tex] from equation (2) to get the DE [tex]\frac{d^2v_x}{dt^2}=-\frac{q^2B_0^2}{m^2}v_x[/tex]

    Here, [tex]\omega =\frac{qB_0}{m}[/tex]

    From this [tex]V_x=Asin(\omega t+\phi)[/tex]....(3)

    Since the initial velocity in the x direction is zero, [tex]\phi[/tex] is zero. Also, since initially velocity is zero, the magnetic force is zero and only electric force is acting.

    Differentiating equation (3) to obtain the acceleration in the x direction.

    [tex]a_x=A\omega cos\omega t=\frac{qE_0}{m}[/tex]
    At t=0, coswt=1, [tex]=\frac{AqB_0}{m}=\frac{qE_0}{m}[/tex]
    This gives you [tex] A=\frac{E_0}{B_0}[/tex]

    From this, you should be able to find everything else. If you cant, I cant help you.
  14. Mar 17, 2007 #13
    They are antiparallel and cancel each other out. You seem to know your calc, but this is an APB style question and all you need are the fundamentals

    The right hand rule tells us that the velocity of the particle, the magnetic field lines, and the force on the particle due to the magnetic field are all mutually perpendicular.

    Place your right fingers in the direction the particle is moving. Now curl your fingers in the direction of the field while keeping your wrist in the same position. If you do this corectly, your thumb should point up, the direction of the magnetic force.

    Once the particle leaves the magnetic field, something interesting happens. The force on the particle is always perpendicular to the velocity as stated before. This should sound familiar as this is the way centripetal force works. The equation in the first post was derived from qvbsin90 = mv^2/r , because the magnetic force is the centripetal force.

    This brings us back to the original question. How fast is it going? In order to be traveling in a straight line in the beginning, the upward magnetic force must be equal in magnitude and opposite in direction to the downward electric force. qvBsin90 = qE. Nothing perpendicular about that
  15. Mar 18, 2007 #14
    I decided to look at the equation again and look at this post again and now I am seriously confuse. I think I am going to find a tutor tomorrow to help me but I was told the physics tutor are not very helpful but I am going to give it a try anyways. BTW teacher emailed me back that she is not going to cover this but I still don't understand why she give us this problem.

    Thanks for all the explanation. (Physics is not for me....)
  16. Mar 21, 2007 #15
    Okay, i finally got to working on this problem again with another classmate and came to this.

    I am not sure if this is correct but this is what i came up with

    Fm = Fe so,

    qvB = qE

    V = E/B

    so I found the constant velocity which is 850/.910 = 934.07 m/s

    after it travels into the reflector that is when the velocity change.

    that is when i thought of cosine and sine but not sure how to do that. im thinking its

    Vcostheta + Vsintheta = Vfinal or cosV + sinV = Vf

    that is how far i got and stop since i didnt want to confuse myself. but if i find the Vf i can then use the r = mv/qB
    Last edited: Mar 21, 2007
  17. Mar 21, 2007 #16
    The particle is accelerating in a circle, but its velocity is only changing direction. Force is perpendicular to velocity, so you can just plug 934.07 m/s in to your equation because its constant
  18. Mar 21, 2007 #17
    Thanks for all the help and explanation and I was finally able to get the answer correctly.
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