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Magnet in copper tube lab calculations (Lenz's Law)

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data
    a magnet that weighs 2.6 grams is dropped through a 1.53 metre tube. the average time taken for it to go through is 1.59 seconds.

    I need to find the strength of the magnetic field and the current


    2. Relevant equations

    Fg=mg
    B=F/Ilsintheta


    3. The attempt at a solution

    Fg=0.025N[down]
    Fsum=0.0031N[down]
    Fm=0.023N[up]

    I=?
    B=?
    l=1.53m
    Fm=0.023N
     
    Last edited: Mar 7, 2010
  2. jcsd
  3. Mar 7, 2010 #2
    any help or guidance would be greatly appreciated.
     
  4. Mar 7, 2010 #3

    lightgrav

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    the magnet speed is essentially constant here, at 1 m/s .
    (avg. accel = 1.2 m/s^2)

    What do you mean by Ft ?
     
  5. Mar 7, 2010 #4
    the sum of the forces or total force
     
  6. Mar 7, 2010 #5

    lightgrav

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    so , where is this current ... do we know its length?

    what causes this current to flow?
     
  7. Mar 7, 2010 #6
    the magnetic field induces a current in the copper tube
     
  8. Mar 7, 2010 #7

    lightgrav

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    so do you think the length of the current is 2pi r , or 1.5 m ?

    and we need the velocity of the copper , relative to the magnet.
     
  9. Mar 7, 2010 #8
    i don't know

    velocity of the copper?
     
  10. Mar 7, 2010 #9

    lightgrav

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    the current travels AROUND the circumference of the tube ; L = 2 pi R .

    velocity of the copper RELATIVE TO the magnet , is velocity of its electrons thru B.

    What direction is the magnetic Field in the copper? DRAW IT
    Force on the electrons = e v x B
     
  11. Mar 7, 2010 #10
    the magnetic field is pointing up
     
  12. Mar 7, 2010 #11
    what about the magnitudes of the current and magnetic field, what equations do i use?
     
  13. Mar 7, 2010 #12

    lightgrav

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    Magnetic Field caused by the bar magnet , within the copper tube,
    has to point OUT through the copper ,
    in order to intercept the charges and apply Force to them.

    As seen by the bar magnet : the electrons in the copper were moving upward ,
    when they got into the magnet's B-field and were deflected by horizontal Force.
    in F = q v x B , F is perp. to v and is perp. to B .
     
  14. Mar 7, 2010 #13
    so i use F=qvB to find B? I don't know the charge either
     
  15. Mar 7, 2010 #14
    how did you find acceleration, i got a different value

    What is the net force
     
  16. Mar 7, 2010 #15

    lightgrav

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    = ½ a t^2 , so a = 2x/t^2 , but that's an "average" acceleration.
    at slow magnet speed, the electrons have very little Force applied to them,
    so they make a very small current around the copper tube
    and waste very little Power (as I^2 R in the tube as resistor, ).
    So the magnet very quickly reaches "terminal velocity" stays at that speed,
    where its grav.PE power = F v is dissipated in the tube's I^2 R = I V .
    and grav. Force mg is cancelled by I 2 pi r B

    so use constant speed, about 1 m/s
     
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