# Conservation of energy in a solenoid

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.

• @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. – Kyle Kanos Mar 23 '17 at 10:03
• @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. – CoilKid Mar 23 '17 at 11:54
• 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. – sammy gerbil Mar 23 '17 at 16:36
• @sammygerbil I updated my answer to include my solution to that conundrum. Hopefully it makes sense! – CoilKid Mar 23 '17 at 17:49

I think I've worked out a solution after sketching some things out.

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 $$W_{bar} = |F_{bar}| \cdot |s| * \text{cos } \theta$$ $$W_{bar} = |-F_{solenoid}| \cdot |L| * \text{cos } 0^{\circ}$$ $$W_{bar} = |\frac{\Delta u_{solenoid}}{L}| \cdot |L|$$ $$W_{bar} = |\Delta u_{solenoid}|$$

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.

• 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. – sammy gerbil Mar 23 '17 at 15:53