# Wire loop falling out of a magnetic field

#### TopherF

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.

2. 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: $v(t)= \frac{g}{\gamma}(1-e^{-\gamma t})$

3. The Attempt at a Solution

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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#### kuruman

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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:

#### TopherF

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

#### haruspex

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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.

#### TopherF

True, but I don't think we need these to answer the question. The unknowns cancel out.
You are correct, they do cancel out.

#### haruspex

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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).

#### TopherF

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

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#### haruspex

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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.

#### TopherF

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.

#### haruspex

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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.
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.

#### TopherF

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 travelling 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 $t_{out}$>$t_{90}$.

#### haruspex

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I know that the loop will be travelling at terminal velocity when it exits
It can never reach terminal velocity.

#### TopherF

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?

#### haruspex

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Will it just approach the terminal velocity in limit?
Yes. That is what your solution to the DE says.
Because there must me a point at which the acceleration due to the induced force is equal to the acceleration of gravity,
Why must?

#### TopherF

Why must?
Assuming that the loop is large enough for it to reach vterm. Otherwise looking back at my DE, you are right

#### haruspex

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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.