Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    Could someone explain how a velocity selector works exactly? It is part of a mass spectrometer, which I can understand, but not quite the velocity selector.

    Have a look at this diagram for example:



    What exactly is B and what does it do? Is it a magnet, that attracts the ion, so that it stays in a straight line and doesn't turn towards the negative terminal (top) ?

    If it is a magnet, does that mean that an electric charge can be attracted/repelled by a magnet as well as another charged particle?
    Offline

    10
    An electric charge is not attracted or repelled by a magnet, but a moving charge is deflected by a magnetic field.
    In the diagram the magnetic field points out of the page and this deflects the moving positive charges (q coulomb moving from left to right) in the downwards direction.
    On the other hand, the electric field is pointing upwards and the lower plate is positive. This deflects the positive charges upwards.
    If you balance the two forces the beam will be undeflected.

    The maths is such that the electric force F_e = Eq
    the magnetic force is F_b = Bqv where B is the magnetic flux density and v the speed of the charges.
    Equating the two forces will give you the expression shown.
    It means that only charges with a velocity v will be undeflected. The selector has a small opening at the end on the right that allows only those undeflected charges through. This means those getting through all have the same velocity and by varying E or B (usually E) you can select what velocity that is.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Stonebridge)
    An electric charge is not attracted or repelled by a magnet, but a moving charge is deflected by a magnetic field.
    In the diagram the magnetic field points out of the page and this deflects the moving positive charges (q coulomb moving from left to right) in the downwards direction.
    On the other hand, the electric field is pointing upwards and the lower plate is positive. This deflects the positive charges upwards.
    If you balance the two forces the beam will be undeflected.

    The maths is such that the electric force F_e = Eq
    the magnetic force is F_b = Bqv where B is the magnetic flux density and v the speed of the charges.
    Equating the two forces will give you the expression shown.
    It means that only charges with a velocity v will be undeflected. The selector has a small opening at the end on the right that allows only those undeflected charges through. This means those getting through all have the same velocity and by varying E or B (usually E) you can select what velocity that is.
    Thanks for your efforts. I know about the maths and all, but just don't understand what B is still.

    Is it a magnetic field then?

    And also: Does gravity play any role in anything in here?

    So, if I understand it correctly, the positive ion gets deflected upwards due to the positive terminal at the bottom. And at a different velocity it would turn downwards, but in that case would it be gravity pulling it downwards or what?

    Thanks
    Offline

    10
    It's nothing to do with gravity. "Down" in the diagram could be any direction.
    B is the "strength" of the magnetic field in Tesla.
    If a charge moves with velocity v in a magnetic field it experiences a force at right angles to its direction of motion. The charge moves at right angles to the field, and the deflecting force is at right angles to both the field and the direction the charge is moving.
    It is summed up in the famous "Left Hand Rule". Below.
    B represents the magnetic field and is in the direction of the pointer finger.
    The middle finger points in the direction of motion of the positive charges and the thumb points in the direction of the deflecting force.
    If you try it for the diagram you posted, you will see that the deflecting force is downwards. (The magnetic field points out of the page)
    (Gravity has a disappearingly tiny effect on a moving charge when you consider how small its mass is.)
    Offline

    0
    ReputationRep:
    (Original post by muffingg)
    And also: Does gravity play any role in anything in here?

    So, if I understand it correctly, the positive ion gets deflected upwards due to the positive terminal at the bottom. And at a different velocity it would turn downwards, but in that case would it be gravity pulling it downwards or what?

    Thanks
    As Stonebridge said, the effect of gravity is negligible. To illustrate it let's compare the orders of magnitude of gravitational and electric forces.

    Electron weighs m_eg=9.11\times 10^{-31}\mathrm{\: kg}\times 9.81\mathrm{\: m/s^2}=8.94\times 10^{-30}\mathrm{\: N}.

    If the electron was in an electric field of 1\mathrm{\: V/m}, a force of eE=1.60\times 10^{-19} \mathrm{\: C}\times 1.00 \mathrm{\: V/m}=1.60\times 10^{-19}\mathrm{\: N} would act on it.

    The ratio of the forces would then be F_e/W=\left( 1.60\times 10^{-19}\mathrm{\: N}\right) /\left( 8.94\times 10^{-30}\mathrm{\: N}\right) \sim 10^{10}. So as you see, neglecting gravitational force is fully justified
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by jaroc)
    As Stonebridge said, the effect of gravity is negligible. To illustrate it let's compare the orders of magnitude of gravitational and electric forces.

    Electron weighs m_eg=9.11\times 10^{-31}\mathrm{\: kg}\times 9.81\mathrm{\: m/s^2}=8.94\times 10^{-30}\mathrm{\: N}.

    If the electron was in an electric field of 1\mathrm{\: V/m}, a force of eE=1.60\times 10^{-19} \mathrm{\: C}\times 1.00 \mathrm{\: V/m}=1.60\times 10^{-19}\mathrm{\: N} would act on it.

    The ratio of the forces would then be F_e/W=\left( 1.60\times 10^{-19}\mathrm{\: N}\right) /\left( 8.94\times 10^{-30}\mathrm{\: N}\right) \sim 10^{10}. So as you see, neglecting gravitational force is fully justified
    Thanks. I get that bit.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Stonebridge)
    It's nothing to do with gravity. "Down" in the diagram could be any direction.
    B is the "strength" of the magnetic field in Tesla.
    If a charge moves with velocity v in a magnetic field it experiences a force at right angles to its direction of motion. The charge moves at right angles to the field, and the deflecting force is at right angles to both the field and the direction the charge is moving.
    It is summed up in the famous "Left Hand Rule". Below.
    B represents the magnetic field and is in the direction of the pointer finger.
    The middle finger points in the direction of motion of the positive charges and the thumb points in the direction of the deflecting force.
    If you try it for the diagram you posted, you will see that the deflecting force is downwards. (The magnetic field points out of the page)
    (Gravity has a disappearingly tiny effect on a moving charge when you consider how small its mass is.)
    So B is actually a straight magnetic field line towards me in the 3rd dimension, right? I understand that if this is the case, using the left hand rule, the motion (thumb) will be downwards. (I hope I get this right this time !?)

    Then in which case would it go towards the top? Does the velocity have to be less or more than v for it to go towards the top?
    Offline

    10
    Yes, correct about the left hand rule.

    So what happens is that the charges coming in from the left will have a range of velocities. (You want to select only those with a value v, shall we say.)
    1. They are all deflected upwards by the electric field with the same force. (F=Eq, it doesn't depend on their velocity, just the strength of the electric field - that would be the applied pd in practice)
    2. They are all deflected downwards by their motion in the magnetic field. The strength of the downwards deflection depends on their velocity, however. (F=Bqv) This means that the faster ones will get a larger downwards force than the slower ones. So the slower ones will tend to deflect upwards because the electric force wins over the magnetic, and the faster ones will tend to be deflected downwards because the magnetic force is greater than the electric force for them. And, of course, some will have just the right velocity that they are not deflected at all because the electric force and the magnetic force are balanced. When Eq=Bqv
    These pass through a small opening in the centre on the right.
    The ones that do this will be those where v=E/B
    All the others will go up or down and miss the opening in the end.
    The result is a beam coming out of the opening on the right where all the charges have the same speed.

    You could adjust the speed you select by adjusting the value of E/B
    You could do this by varying the pd across the plates, as this varies E.
    A higher pd (higher E) would select charges with a higher speed.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Stonebridge)
    Yes, correct about the left hand rule.

    So what happens is that the charges coming in from the left will have a range of velocities. (You want to select only those with a value v, shall we say.)
    1. They are all deflected upwards by the electric field with the same force. (F=Eq, it doesn't depend on their velocity, just the strength of the electric field - that would be the applied pd in practice)
    2. They are all deflected downwards by their motion in the magnetic field. The strength of the downwards deflection depends on their velocity, however. (F=Bqv) This means that the faster ones will get a larger downwards force than the slower ones. So the slower ones will tend to deflect upwards because the electric force wins over the magnetic, and the faster ones will tend to be deflected downwards because the magnetic force is greater than the electric force for them. And, of course, some will have just the right velocity that they are not deflected at all because the electric force and the magnetic force are balanced. When Eq=Bqv
    These pass through a small opening in the centre on the right.
    The ones that do this will be those where v=E/B
    All the others will go up or down and miss the opening in the end.
    The result is a beam coming out of the opening on the right where all the charges have the same speed.

    You could adjust the speed you select by adjusting the value of E/B
    You could do this by varying the pd across the plates, as this varies E.
    A higher pd (higher E) would select charges with a higher speed.
    Thanks a lot again. Shame I cant give you another pos rep. That last comment of yours finally explained it all, which my teacher never managed to explain to anyone.

    Great help.
    Offline

    9
    ReputationRep:
    and it doesnt matter what the masses or charges are (even + or -) . What emerge have the same speed.
    Offline

    10
    (Original post by teachercol)
    and it doesnt matter what the masses or charges are (even + or -) . What emerge have the same speed.
    Yes. An important point.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 23, 2011

University open days

  1. Loughborough University
    General Open Day Undergraduate
    Fri, 21 Sep '18
  2. University of Cambridge
    Churchill College Undergraduate
    Fri, 21 Sep '18
  3. Richmond, The American International University in London
    Undergraduate Open Day Undergraduate
    Fri, 21 Sep '18
Poll
Which accompaniment is best?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.