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Lorentz Force and right hand rule

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    > Figure 1: XWSl6.png
    >
    > A spool is hung on a spring. A part of the coil hangs in front of a
    > homogeneous magnetic field. The coil is incorporated in an electrical
    > circuit. See Figure 1.
    >
    > the coil KLMN has a rectangle shape, has a height of 10.0 cm, a width
    > of 8.0 cm, and has 200 windings of which two are shown in Figure 1.
    >
    > The electrical circuit comprises a power supply, a sliding resistance
    > and the coil, connected in the manner shown in Figure 1
    >
    >
    > The distance between the bottom of the coil and the borders of the
    > magnetic field is referred to as y. See Figure 1. If the power source
    > is not yet connected, y is equal to 5.0 cm. The voltage source will
    > now be connected, so that the spring will be stretched further and a
    > Lorentz force will be exerted. The current in the coil can be changed
    > by chaging the resistance. If there is a larger current, the spring
    > will stretch more.

    A.When calculating the LorentzForce : Fl=N*B*I*l Is the length (l) 8cm or 10cm? And why?

    B. If the value of the current intensity is at a certain value, the coil length does not appear to increase any further with an increase in the current intensity.

    Why is this the case when y=10 cm and how to apply the righthandrule in here?






    2. Relevant equations
    Fl=N*B*I*l
    righthandrule



    3. The attempt at a solution
    A. I don't know why it should be 8cm...

    I think 10cm, is because the height of the coil. But I don't see why it should be ten, because when there is no current y is already 5. And I don't see how you can use the righthandrule to show that the coil is going upward after y=10cm.
     
  2. jcsd
  3. Nov 11, 2013 #2
    The current is up in the KL section of the coil, down in the MN section and from left to right in the NK section.
     
  4. Nov 11, 2013 #3

    rude man

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    What is the direction of the force on the horizontal and vertical sections of the coil?

    The coil does not go upward at y = 10 cm. It stays at 10 cm.

    BTW it's not stated but from the description of the situation it's obvious that the B field points out of the page.
     
  5. Nov 12, 2013 #4

    tiny-tim

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    Hi Edira! Welcome to PF! :smile:

    I'll just add …

    it's Laplace force, not Lorentz force! :wink:
     
  6. Nov 12, 2013 #5

    vanhees71

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    Why do you say it's Laplace Force? I've never heard about a Laplace force.

    The Lorentz force, however, is indeed the well-established force acting on a charged particle in the electromagnetic field (written in Gaussian or Heaviside-Lorentz CGS units),
    [tex]\vec{F}=q \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]
    The continuum-mechanics version is the force density (force per volume fluid),
    [tex]\vec{f}=\rho \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right ) = \rho \vec{E} + \frac{1}{c} \vec{j} \times \vec{B},[/tex]
    where [itex]\rho[/itex] is the charge density and [itex]\vec{j}[/itex] the current density of the charged fluid.

    In your case you can get the total Lorentz force on the spool by
    [tex]\vec{F}_{\text{mag}}=\frac{N I}{c} \int_{\text{spool}} \mathrm{d} \vec{r} \times \vec{B},[/tex]
    because for simplicity you can model the spool as an infinitely thin wire wound [itex]N[/itex] times over the rectangular frame.

    I guess, what you are supposed to calculate is the situation where the spool is at rest (the fully dynamical problem is not so simple, particularly when you want to include the induced current and the backreaction to the motion of hte spool). For the static case you have to evaluate the force balance
    [tex]\vec{F}_{\text{grav.}} +\vec{F}_{\text{spring}} + \vec{F}_{\text{mag}}=0[/tex]
    from which you can determine the value [itex]y[/itex] for a given current.

    Note that there is a change as soon as the spool is fully within the region of the magnetic field!
     
  7. Nov 12, 2013 #6

    rude man

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    ??? It's the Lorentz force.
     
  8. Nov 14, 2013 #7
    Isn't Lorentz force 'Fnet= qE + q(V x B) = q{E + (V x B)}'?
     
  9. Nov 14, 2013 #8

    rude man

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    Definitely!
     
  10. Nov 15, 2013 #9

    tiny-tim

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    see eg http://en.wikipedia.org/wiki/Lorentz_force#Force_on_a_current-carrying_wire

    the lorentz force is the name usually given to the force on a charged body …
    it seems to me unhelpful to give the same name to the force on an uncharged body (a wire) in which the positive charges are stationary and the equal negative charges are moving :wink:
     
  11. Nov 15, 2013 #10

    rude man

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    The force on a moving charge (aka "current") in a B field is the same whether the charge is moving in space or in a wire.

    The force is not on the wire per se, it's on the moving charges. Strong binding forces convey the force on the charges to a force on the wire. :wink:
     
  12. Nov 15, 2013 #11

    tiny-tim

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    but the lorentz force contains a E component

    if the charge is not in a neutral wire, the lorentz force will not be NBIL :wink:
     
  13. Nov 15, 2013 #12

    rude man

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    The Lorentz force is a general expression encompassing moving charge in a B field and static or moving charge in an E field.

    Actually, there is also an E field and forces on static charges. The potential differences between different sections of the coil imply a corresponding E field between those sections and therefore free charges along the coil. So the entire Lorentz expression applies.

    The E forces are in this case of negligible impact. :wink:
     
  14. Nov 16, 2013 #13

    tiny-tim

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    but that's not an external E field

    the lorentz force contains a E component from any external E field

    if the charge is not in a neutral wire, the lorentz force will not be NBIL
     
  15. Nov 16, 2013 #14

    rude man

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    A surface charge doesn't know that! All it sees is an E field and a force given by the Lorentz expression for an E field. :wink:
     
  16. Nov 17, 2013 #15

    tiny-tim

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    not following you :confused:
     
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