Wire loop falling out of a magnetic field

In summary, the loop will leave the field at velocity .012 m/s when the y-axis reaches the value of .
  • #1
TopherF
8
1

Homework Statement



Here is a link to the problem: https://imgur.com/a/0T9sF

I am having trouble finding the time it takes the loop to leave the field in part a.

Homework Equations



## emf=-\frac{d \Phi}{dt} ##

## \Sigma F=ma=mg-BIl ##

## \frac{dv}{dt}-\gamma v=g ## where ## \gamma=826.72##.

Solution to that DE: ## v(t)= \frac{g}{\gamma}(1-e^{-\gamma t}) ##

The Attempt at a Solution


[/B]
In part a, I determined that ##v_{term}=\frac{g}{\gamma}=.012## m/s, and that ##t_{90}=.0028## s.

I used separation of variables on the differential equation and integrated to get position as a function of time:

## y-y_{o}=\frac{1}{2}(g-\gamma v)t^2 ##

I tried letting ##y_{o}=0## and ##y_{o}=l## to solve for t. But I am not given ##l##. My professor claims I do not need to know it but I am having a hard time seeing how I can find the time at which the loop leaves the field without it. I also integrated ##v(t)## to get position as a function of time but I am running into the same probloem.
 
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  • #2
It doesn't look like you have found the correct equation of motion. ##ma=mg-BIl## is a good start, but what did you do with the induced current I? What is the relation of ##\gamma## to the given quantities and where did 826.72 come from?

On edit: I noticed that the diameter (gauge) of the wire is not given. Without it you cannot find the mass of the loop or its resistance.
 
Last edited:
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  • #3
I calculated the current by ## I=\frac{emf}{R} ## and the resistance by ##R=\frac{\rho_{e} (4l)}{A}##

I then calculated the emf by taking the negative time derivative of the flux. Using the emf and the resistance I got ##I##.

My professor claims this problem will be independent of mass.

This is my work so far: https://imgur.com/a/ZMNUu
 
  • #4
kuruman said:
Without it you cannot find the mass of the loop or its resistance.
True, but I don't think we need these to answer the question. The unknowns cancel out.
 
  • #5
haruspex said:
True, but I don't think we need these to answer the question. The unknowns cancel out.

You are correct, they do cancel out.
 
  • #6
TopherF said:
You are correct, they do cancel out.
I think I agree with your formula for vterm (it is hard to read) but I get a somewhat smaller number (2/3 of yours).
 
  • #7
haruspex said:
I think I agree with your formula for vterm (it is hard to read) but I get a somewhat smaller number (2/3 of yours).

I got that ##v_{term}=\frac{g}{\gamma}=\frac{16g \rho_{m} \rho_{e}}{B^2}=\frac{(16)(9.81)(2700)(2.8*10^{-8})}{(1^{2})}=.0118 ## m/s
 
  • #8
TopherF said:
I got (...) .0118 m/s
So did I.
 
  • #9
TopherF said:
I got that ##v_{term}=\frac{g}{\gamma}=\frac{16g \rho_{m} \rho_{e}}{B^2}=\frac{(16)(9.81)(2700)(2.8*10^{-8})}{(1^{2})}=.0118 ## m/s
Ah yes, my mistake.
 
  • #10
Using the differential equation

## \frac{dv}{dt}=g-\gamma v##

I used separation of variables and integrated from 0 to ##t## to get

## v-v_{o}=(g-\gamma v)t ## and knowing ##v_{o}=0## i got the equation

##\frac{dy}{dt}=(g-\gamma v)t## so separating variables and integrating again I get that

##y-y_{o}=\frac{1}{2}(g-\gamma v)t^{2}##

Positive ##y## is taken to be downwards so ##y_{o}=0## and I need to solve for ##t## when the loop leaves the field, i.e. when ##y=l##

But without knowing ##l## I am not sure where to go from here or if this is the correct approach.I also tried integrating ##\frac{dy}{dt}=\frac{g}{\gamma}(1-e^{-\gamma t})## and I get that

##y=\frac{g}{\gamma^{2}}(e^{-\gamma t}+\gamma t-1) ## but I end up in the same boat.
 
  • #11
TopherF said:
Using the differential equation

## \frac{dv}{dt}=g-\gamma v##

I used separation of variables and integrated from 0 to ##t## to get

## v-v_{o}=(g-\gamma v)t ## and knowing ##v_{o}=0## i got the equation

##\frac{dy}{dt}=(g-\gamma v)t## so separating variables and integrating again I get that

##y-y_{o}=\frac{1}{2}(g-\gamma v)t^{2}##
You cannot integrate like that. v is a function of time.
TopherF said:
I also tried integrating ##\frac{dy}{dt}=\frac{g}{\gamma}(1-e^{-\gamma t})## and I get that

##y=\frac{g}{\gamma^{2}}(e^{-\gamma t}+\gamma t-1) ## but I end up in the same boat.
That is the right way, but I agree there is no way to get a numeric answer without knowing L.
This can be easily seen by setting the field to zero.
 
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  • #12
haruspex said:
You cannot integrate like that. v is a function of time.
That is the right way, but I agree that is no way to get a numeric answer without knowing L.
This can be easily seen by setting the field to zero.

Okay, I overlooked that. Thanks.

I think I may have figured something out. Would you tell me if mathematically this is legal and this process makes sense to you in the context of this problem?

I took the bottom equation from my previous reply and substituted ##y=vt##. Then, I know that the loop will be traveling at terminal velocity when it exits so I plugged in ##v=.012## m/s. I then divided the entire equation by ##t## to get that:

##v_{term}=\frac{\frac{g}{\gamma^2}e^{-\gamma t}+.012t-\frac{g}{\gamma^2}}{t}##

I then plotted this function and just graphically found the ##t## value when the value of the function is ##0.012##.

I found that ##t_{out}=.0108## s. This has some physical sense because ##t_{out}##>##t_{90}##.
 
  • #13
TopherF said:
I know that the loop will be traveling at terminal velocity when it exits
It can never reach terminal velocity.
 
  • #14
haruspex said:
It can never reach terminal velocity.

Could you explain this a little more? Will it just approach the terminal velocity in limit? Because there must me a point at which the acceleration due to the induced force is equal to the acceleration of gravity, so the net acceleration is zero, so it must be moving at some constant speed, even if for a brief time?
 
  • #15
TopherF said:
Will it just approach the terminal velocity in limit?
Yes. That is what your solution to the DE says.
TopherF said:
Because there must me a point at which the acceleration due to the induced force is equal to the acceleration of gravity,
Why must?
 
  • #16
haruspex said:
Why must?

Assuming that the loop is large enough for it to reach vterm. Otherwise looking back at my DE, you are right
 
  • #17
TopherF said:
Assuming that the loop is large enough for it to reach vterm
But it can never be.
I note the next part of the question where it asks about 90% of terminal velocity, which maybe it can reach.
 

What causes a wire loop to fall out of a magnetic field?

The force of gravity acting on the wire loop causes it to fall out of a magnetic field.

How does a magnetic field interact with a wire loop?

A magnetic field exerts a force on a wire loop, causing it to move in a circular motion.

Why does a wire loop fall at a constant speed out of a magnetic field?

The wire loop falls at a constant speed because the force of gravity is equal to the force exerted by the magnetic field, resulting in balanced forces.

Can a wire loop fall out of a magnetic field at different speeds?

Yes, the speed at which a wire loop falls out of a magnetic field can vary depending on factors such as the strength of the magnetic field and the weight of the wire loop.

How does the size of a wire loop affect its interaction with a magnetic field?

The size of a wire loop does not affect its interaction with a magnetic field. The force exerted by the magnetic field is dependent on the strength of the field and the current flowing through the wire, not the size of the loop.

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