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Magnets down a copper tube and Lenz's law.

  1. Dec 8, 2009 #1
    Right. I know if you drop a strong magnet down a copper tube (or conductive but non magnetic tube) it falls slowly as it 'sees' a changing magnetic field and thus induces a current in the tube and this induced current causes its own magnetic field to be induced and acts a resistive force to the falling magnet.

    But i've been trying to work out how the number of magnets (dropped as one body) is related to the velocity at which it falls. Could anyone help me with this one? I've been trying to work it out for ages but can't seem to get very far with it.

    Thanks in advance for any help!
     
  2. jcsd
  3. Dec 9, 2009 #2
    I’ll see if I can help.
    Say your copper tube has an inside radius R and wall thickness D. Suppose R>>D. Say the (average) length of the magnetic field lines outside the magnet is H. Strength of magnetic field is B.

    Then starting with the upwards force F on the magnet exerted by the currents I flowing in the tube:

    F=BIL (I don’t know the name of this formula).

    Induced currents I are running in 2 circles along the circumference inside the tube. One clockwise say for the north pole and ccw for the south pole, depending on which way the magnet is turned. The current density J is I/A where area A=D x H.
    Then I=JxDxH

    L is along the direction of the current so here: L =2pi R.

    Put this back in F. F=BxJxDxHx2pixR. Now call DxHx2piXR=Volume=Vol.
    This is the effective volume of copper occupied by B.
    So: F=BJVol
    Now J=E/ρ where rho is resistivity of copper and E is the electric field generated by BxV of the falling magnet. J=BV/ρ

    Hence: F=B^2VVol/ρ. This the force per magnet. If you bundle some N magnets close together then H is the length of each magnet and F is multiplied by N. Also factor in an efficiency in of perhaps 70 to 80% because of field lines running in the air space between magnet and tube.

    The electrical power P generated in the tube is: P=FxV=(B^2V^2Vol/ρ)x%. You can check this out because P is also I^2 x resistance.
    The resistance is 2pixRxρ/HD.

    I’ve never done this experiment so let us know if my calculations make any sense.
     
  4. Dec 10, 2009 #3
    http://www.youtube.com/watch?v=30oPZO_z7-4&feature=related

    This film confirms my calculations. Those small magnets can have magnetic B fields of up to 0.66 Tesla and with say a weight of 50 gram. If you put all that in the formula then V is very small say ~5 cm per sec.
     
  5. Dec 17, 2009 #4
    Per Oni, thanks for you help. Your method makes sense but is not conclusive with my data. I think it may be easier if you see my data than me describing what i've found!

    N V
    1 0.568
    2 0.279
    3 0.269
    4 0.264
    5 0.318
    6 0.370

    N is the number of magnets, V is the average velocity (which equals terminal velocity - acceleration is negligible).

    I think the relationship is to do with the changing shape of the magnetic field - with 3 or 4 magnets more of the field lines are perpendicular to the tube and thus a greater current is induced so a minimum terminal velocity is reached. What do you think?
     
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