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I found a website this afternoon which stated that for a given solenoid, the force it exerts on an iron bar can be modeled as a conservation of energy problem.

The page gives a method to find $\Delta u_{solenoid}$. It then says that the force exerted on the bar can be modeled as

$$F= \frac{\Delta u_{solenoid}}{\text{length of solenoid}}$$

This is a problem for me because I need the work done on the bar.

For a bar of identical length to the solenoid, if I try to utilize \begin{align} W &= |F|\cdot|s| *\text{cos }\theta\\ &= |\frac{\Delta u_{solenoid}}{\text{length of solenoid}}| \cdot|\text{length of solenoid}| *\text{cos }\theta \\ &=|\Delta u_{solenoid}|*\text{cos }\theta \end{align}

Then I'm stuck with something dependent on $\theta$. Intuitively, and experimentally, I know that magnets pull on iron which suggests to me that in this scenario $\theta$ is always $0$. (Both poles of a magnet pull iron, etc.)

However, this doesn't seem like the "proper" way to do it, and I was wondering if anyone could explain analytically what $\theta$ has to be and why.

Or, considering the not-improbable scenario in which I'm approaching this problem all wrong, what would be the correct way to get $W$ from my $\Delta u_{solenoid}$?

Edit:

I should have explained a bit better. The way that I view this

$$ 0=\Delta KE + \Delta u$$ $$\Delta u = -\Delta KE$$

Which suggests to me that when the iron bar enters the solenoid, $KE$ is reduced and negative work is done on the bar.

Experimentally, this isn't the case, however, and I'm trying to determine what's going on. I suspect that it has to do with a change in potential energy of my power source, but I'm not familiar enough with electrical and magnetic potentials to say.

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    $\begingroup$ @sammygerbil: what do you mean it isn't clear why $\theta$ was introduced? $W=Fd\cos\theta$ is a common definition for work (for a constant force not in the direction of $d$), it even appears on Wikipedia's entry. $\endgroup$ – Kyle Kanos Mar 23 '17 at 10:03
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    $\begingroup$ @sammygerbil I don't see how you can get a positive work on the iron bar if $\Delta u_{solenoid}$ is also positive. I'll try and explain in my question a bit better. $\endgroup$ – CoilKid Mar 23 '17 at 11:54
  • $\begingroup$ I agree with you that an increase in magnetic energy while work is being done seems to be against the conservation of energy. There is a real puzzle here which requires an explanation. So I have retracted my close vote and given you an upvote. $\endgroup$ – sammy gerbil Mar 23 '17 at 16:36
  • $\begingroup$ @sammygerbil I updated my answer to include my solution to that conundrum. Hopefully it makes sense! $\endgroup$ – CoilKid Mar 23 '17 at 17:49
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I think I've worked out a solution after sketching some things out.

enter image description here

If we take the iron bar to be motionless and the solenoid to be moving, then the force on the solenoid is $$F_{solenoid}=\frac{\Delta u_{solenoid}}{-L}$$ From Newton's Third Law, we know that there's an antiparallel force of the same magnitude acting on the bar. $$F_{solenoid}=-F_{bar}$$

Now if we consider the solenoid to be stationary and the iron bar to be in motion, then $$W_{bar} = |F_{bar}| \cdot |s| * \text{cos } \theta$$ $$W_{bar} = |-F_{solenoid}| \cdot |L| * \text{cos } 0^{\circ}$$ $$W_{bar} = |\frac{\Delta u_{solenoid}}{L}| \cdot |L|$$ $$W_{bar} = |\Delta u_{solenoid}|$$

Which answers my question.

I think I was accidentally confusing everything by treating the $KE$ of my solenoid as the $KE$ of the iron bar.

Edit:

To satisfy Conservation of Energy:

If one has an inductor, there must be a power source somewhere with $E=u_{battery}$. For the solenoid/battery system, which is stationary if the bar is moving:

In general: $$W_{other}=\Delta KE_{solenoid} + \Delta u$$ $$-W_{bar} - W_{heating} = \Delta KE_{solenoid} + \Delta u_{field} + \Delta u_{battery}$$ $$-W_{bar} - W_{heating} = (0) + \Delta u_{field} + \Delta u_{battery}$$ and specifically for our situation: $$-|\Delta u_{field}| - W_{heating} =\Delta u_{field} + \Delta u_{battery}$$ $$-\Delta u_{field} -|\Delta u_{field}| - W_{heating} =\Delta u_{battery}$$ And because our $\Delta u_{field}>0$ $$-2\Delta u_{field} - W_{heating} =\Delta u_{battery}$$ $$\Delta u_{battery}= -2\Delta u_{field} - W_{heating} $$

Which shows that our example does indeed follow the Law of Conservation of Energy. We're not getting energy from nothing; it must come from the power source.

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  • $\begingroup$ Your final result is just the same as what the website says. But I am not convinced that you have explained how energy is conserved. The bar will also gain KE. ... A dielectric slab being drawn into a parallel plate capacitor is similar. $\endgroup$ – sammy gerbil Mar 23 '17 at 15:53

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