# Equation for F using flux desnity in solenoid

1. Aug 3, 2009

### Jake110

i've been looking for an equation to find the force exerted on a ball bearing in a solenoid (on axis) using the magnetic flux density but i have no idea where to look, ive googled it, ive tried online physics enciclopedias but i just can't find it. im going through town today and im going to stop off at the libary but if you guys know anything, please post it. any help would be very much appreciated.

to get the flux density im using http://img229.imageshack.us/img229/5841/equation1solenoidimg.png [Broken]

EDIT: Nope they dont have anything. i read through every physics book they had in the science section and didn't find anything i didn't already know. (most of its just the same thing in 3-4 books). the guy at the help desk said its a bit to specialised. does nyone know an equation i can use? or know of a book that has it in?

Last edited by a moderator: May 4, 2017
2. Aug 3, 2009

### SShermer

Jake, could you please possibly post the source for the equation in your previous post, and a list of the variable definitions? This would be helpful for people who can't get access to the text where you found it.

With regards to the above equation, what is the variable x?

Scot

Last edited: Aug 3, 2009
3. Aug 3, 2009

### Jake110

its from http://www.netdenizen.com/emagnettest/solenoids/?solenoid". its all there, plus a diagram. im just trying to find the acceleration the solenoid will give a ballbearing (its going to start outside the solenoid, on axis), i know how to get the acceleration from the force, i just need to find a way to work out the force.

if you want the values im using, look at my https://www.physicsforums.com/showthread.php?t=327325". the only reason i made this new topic was because no one answered my last post for about a week. kind of annoying when i want to get this thing build and tested before my college course starts in september.

Last edited by a moderator: Apr 24, 2017
4. Aug 3, 2009

### gabbagabbahey

To find the force exerted on any object by any electromagnetic field, you simply turn to the Lorentz Force Law...a small piece of any object can be treated as being essentially a point charge $dq=\rho dV$ (where $\rho$ is the localized charge density in the infinitesimal piece of the object $dV$) experiencing an infinitesimal force due to the electric and magnetic fields ($\textbf{E}$ and $\textbf{B}$ respectively) in which it is placed: $d\textbf{F}=\rho(\textbf{E}+\textbf{v}\times\textbf{B})dV$. Hence the entire object experiences a force:

$$\textbf{F}=\int\rho(\textbf{E}+\textbf{v}\times\textbf{B})dV$$
where the (triple) integration is over the volume of the object.

Now, as you should see clearly, if the ball bearing has a charge density of zero, there will be no force on it.

So, the first question you need to ask yourself is whether or not there is any charge density on the ball bearing and why... the main factors that will affect this are (1)what material the ball bearing is made of---is there any ferromagnetic material in it such as iron or cobalt? Is it a conductor or an insulator? and (2)whether there is any electric field present that would induce a charge density on the ball bearing---if you vary the current in the solenoid, you will create a time-varying magnetic field which will induce an electric field (Faraday's Law) which will then induce a non-zero charge density in a conducting ball bearing

5. Aug 3, 2009

Staff Emeritus
gabbagabbahey, that's a highly non-optimal way of solving the problem. Why does a piece of iron get attracted to a magnet? How would you calculate this from the Lorentz force?

It's better to realize that there is a force due to the gradient of a magnetic field,

$$F = \mu \frac{dH}{dz}$$

and solve it that way. (I am not 100% sure I haven't dropped a $4 \pi$).

6. Aug 3, 2009

### Bob S

The Lorentz force law applies only to charged particles or objects moving in a magnetic field. Your ball bearing is neither charged nor moving so forget it. Now, in a magnetic field that has a gradient, like your solenoid, there is a force F on the ball bearing. If the ball bearing is displaced a distance dx in the magnetic field, work is done against this force (See Vanadium 50's post).

Work = W = F dx

This work adds to the total stored energy E in the magnetic field, that can be expereesed either by the engineers equation for inductance and current in the solenoid

E = (1/2)L I2, or by the physicist's equation

E = (1/2) ∫ E H dV = (1/2 u0 ∫ B2 dV (in air)

In either case, dE/dx is the force along a displacement in the x direction.

This is hard to calculate theoretically, but estimates can be made using perturbation approximations.

The ball bearing modifies the stored energy in the field of your solenoid, so by using your equation for the on-axis of the solenoid, the B field is increased dramatically inside the ball bearing because the magnetic field B in steel is about 5000 (relative permeability) times the B in air. This change in extra stored energy depends on x1 and x2 in your equation, so E increases, and there has to be a force. [There is an extra factor due to the external dipole moment of the magnetic dipole induced in the ball bearing, but I forget it]. Calculate the change in stored energy for the ball bearing in two positions differing by a distance dx, and divide the calculated stored energy by dx.

∂ ∫ ∏ ∑

Last edited: Aug 3, 2009
7. Aug 3, 2009

### gabbagabbahey

No, the Lorentz force law applies to all objects and is the most general force law in classical electrodynamics. All other force laws in classical electrodynamics are simply special cases of it.

true enough, but still valid.
The applied field produces a torque (via the lorentz force law) on the molecular/atomic magnetic dipoles in the iron tending to align a certain percentage of them (more precisely, it aligns some of the domains that contain a bunch of dipoles pointing in any one direction) with the field, thus producing a bulk magnetization. Each of these tiny molecular/atomic dipoles can be modeled as tiny current loops, and the force on each current loop can be calculated via the Lorentz formula, and these infinitesimal forces can be suitably averaged (using knowledge of the magnetization or auxiliary field H of the iron piece) ....this is of course, (more or less) how the gradient equation you posted is derived.

Also, if one rapidly varies the current in the solenoid, an electric field will be induced and there will be an additional force (which can be many orders of magnitude stronger than the magnetic force) on a conducting object.

8. Aug 3, 2009

### Bob S

Even when uncharged objects are not moving in a magnetic field?

Even when an uncharged object is in a magnetic field? Remember that the force has to change direction when the magnetic field B is reversed, or it is not the Lorentz v x B force.

The term "rapidly varying" was not in the OP. The largest force on a conducting object in a rapidly varying magnetic field is usually the repulsive eddy current force.

9. Aug 3, 2009

### gabbagabbahey

Yes, why would you think otherwise?

For example, read through Jackson's Classical electrodynamics and you will see that the starting point for the derivation of every specialized torque and force equation is always the same....the Lorentz force Law.

Again, yes.

If you reverse the field, you also tend to flip the magnetic domains so that the currents $\rho\textbf{v}$ also change sign (loosely speaking) and the cross product is unaffected.

10. Aug 3, 2009

### Bob S

from Bob S
[Is there a Lorentz force on an object] Even when uncharged objects are not moving in a magnetic field?
Jackson on pg 572 2nd Edition starts with dp/dt = e( E + (1/c) v x B). Is this the one you mean? This is for the Lorentz force on a moving charged object. The ball bearing in the PO is neither charged nor moving.

From Bob S
Even when an uncharged object is in a magnetic field? Remember that the force has to change direction when the magnetic field B is reversed, or it is not the Lorentz v x B force.
The Lorentz force is only for charged objects, even when you do Lorentz transformations to a moving frame..

The ball bearing in the OP is not charged and is not moving.

11. Aug 4, 2009

Staff Emeritus
But solving it this way is a little like using QM and GR to calculate a block sliding down an inclined plane. It's possible, but it's really not the way to go about solving the problem.

Rather than first principles, it's better to think about E&M in media. Let me see if I can derive the force and get all the constants right.

The magnetic field (actually the magnetic flux density) inside a soft iron sphere in a uniform magnetic field B is 3B. (The 3 is totally irrelevant, but it's not obvious why until the next step) So the energy density inside the sphere is $(3B)^2/2\mu$ and outside the sphere $B^2/2\mu_0$. For soft iron $\mu >> \mu_0$, so we can essentially treat the magnetic field energy inside the sphere as 0. So the energy change in the system due to the sphere is given by:

$$U = -\frac{B^2}{2\mu_0}$$

So

$$F = -\nabla U = \frac{B}{\mu_0}\frac{dB}{dz}$$

which can be written several different ways. It looks like it has the right properties - the force is towards the direction of increasing magnetic field, it's proportional to the magnetic field, and it's proportional to the gradient.

12. Aug 4, 2009

### gabbagabbahey

A bit of an exaggeration, but yes. Anyways, I wasn't really trying to suggest the OP solve the problem directly from the Lorentz force Law, but rather that he first think about why there is any force at all before blindly applying formulas to calculate its magnitude. The why is usually easiest to see by looking at the Lorentz Force Law, and thinking about the properties of the ball bearing and the applied field(s).

That's another version of the Lorentz Force Law, yes. However it applies to a point-like particle of charge $e$, not any charged object as you seem to suggest.

I think I see now the source of your confusion: you seem to be thinking of an uncharged object as necessarily having a charge density of zero; but that simply isn't the case. Neutral matter consists of large numbers of electrons and protons which are charged and occupy different points in space--- at one point in space, you may be very near an electron while more distant from the positive nucleus it is bound to, and hence at that point in space the charge density would be slightly negative. Nearer to the nucleus, the charge density would be positive, but it isn't zero throughout the object. The overall net charge on an object may be zero, but unless it is made of uncharged particles (such as a soup of neutrinos). it will have a nonzero charge density that varies over space....the fact that the object as a whole is neutral only means that $\int\rho dV=0$.

Last edited: Aug 4, 2009
13. Aug 4, 2009

### Jake110

sorry i haven't replied for a few days, my modem is faulty. so if i use this formula i can get the pulling force the solenoid will exert on the ball bearing?

i've only understood about half of what you guys have said, could you give me a list of what all the symbols mean? i know B is the flux desnity and u0 is the permeability of the ball bearing but i dont know what z is and is the gradient worked out the same way as dy/dx?

just to simplify the writing of the equation is it this-

F = (B/u0)(dB/dz)

thats what it looks like to me.

14. Aug 6, 2009

### Jake110

oh man this always happens, i almost get it solved then people stop replying. what is z? and what is the gradient you guys talk? its of the flux density and this z value.

15. Aug 6, 2009

Staff Emeritus
Jake, you have no right to demand answers - or for that matter, to demand anything. What gives you the right to complain that you aren't getting your answers fast enough?

z is x in your notation, which is non-standard. Usually z is the length along the solenoid. As for gradient, it's in the Library - all you have to do is click on the word. How much spoon-feeding do you need?

16. Aug 6, 2009

### Jake110

Well how is it my fault when people just ignore my topics? This is the fifth one I’ve made to try and get this sorted. What the hell is with the sudden hostility anyway? I wasn't demanding anything, I was just stating how freaking annoying it is when you've almost worked something out and then everyone stops talking, how would you react if you started talking with someone and then they suddenly walked off half way through the conversation.

17. Aug 6, 2009

### Bob S

Shouldn't the force on the ball bearing be proportional to the volume of the ball bearing, like V = 4/3 pi R3 ?

18. Aug 7, 2009