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

  • #1

Homework Statement


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


Homework Equations



Fg=mg
B=F/Ilsintheta


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:

Answers and Replies

  • #2
any help or guidance would be greatly appreciated.
 
  • #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 ?
 
  • #4
the sum of the forces or total force
 
  • #5
lightgrav
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so , where is this current ... do we know its length?

what causes this current to flow?
 
  • #6
the magnetic field induces a current in the copper tube
 
  • #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.
 
  • #8
i don't know

velocity of the copper?
 
  • #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
 
  • #10
the magnetic field is pointing up
 
  • #11
what about the magnitudes of the current and magnetic field, what equations do i use?
 
  • #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 .
 
  • #13
so i use F=qvB to find B? I don't know the charge either
 
  • #14
the magnet speed is essentially constant here, at 1 m/s .
(avg. accel = 1.2 m/s^2)

What do you mean by Ft ?

how did you find acceleration, i got a different value

What is the net force
 
  • #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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