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

AI Thread Summary
A 2.6-gram magnet is dropped through a 1.53-meter copper tube, taking 1.59 seconds to pass through, indicating a constant speed of approximately 1 m/s. The discussion centers on calculating the magnetic field strength and induced current, with participants debating the relevant equations and the nature of the forces involved. The magnetic field induces a current in the copper tube, and the forces acting on the electrons are analyzed using Lenz's Law and the equation F = qvB. The net force and average acceleration of the magnet are also discussed, highlighting that the magnet reaches terminal velocity where gravitational potential energy is balanced by the power dissipated in the tube. Understanding the relationship between the magnetic field, current, and forces is crucial for solving the problem.
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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:
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any help or guidance would be greatly appreciated.
 
the magnet speed is essentially constant here, at 1 m/s .
(avg. accel = 1.2 m/s^2)

What do you mean by Ft ?
 
the sum of the forces or total force
 
so , where is this current ... do we know its length?

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

velocity of the copper?
 
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
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
lightgrav said:
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
= 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 canceled by I 2 pi r B

so use constant speed, about 1 m/s
 
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