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Outward force on current loop (Ampere's Force Law)

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Philip Wood
Jan8-13, 04:49 PM
PF Gold
P: 962
I said (post 16) that I thought the force per unit length on the circular loop due to the loop's own current might be simply related to its inductance. Here's a suspiciously simple argument...

Let the magnetic field due to the whole loop at a point on the loop itself be B. This field is directed normally to the plane of the loop.

Suppose we increase the loop radius by dr, while keeping the current constant. [This would be the case if the loop were connected to a high voltage d.c. supply via a high resistance; the back-emf created when the ring's radius is increased would be negligible compared with the supply voltage.] The extra flux d[itex]\Phi[/itex] linked with the loop due to its expansion will be d[itex]\Phi = 2\pi r dr B[/itex]. This is just multiplying the extra loop area by the flux density at the periphery.

But, L is defined (if there are no non-linear materials present) by [itex]\Phi = LI[/itex]. [Differentiate both sides wrt time (at constant L) to get the more familiar form.] So, substituting LI for [itex]\Phi[/itex] in the previous equation, and remembering the constant current, we get
I d[itex]L = 2 \pi B r dr[/itex].

That is [itex]B = \frac{I}{2 \pi r}\frac{dL}{dr} [/itex].

But the outward force per unit length of circumference is [itex]\frac{F}{s} = BI[/itex].

So [itex]\frac{F}{s} = \frac{I^2}{2 \pi r}\frac{dL}{dr} [/itex].

Knowing L as a function of r, we can find the force per unit length on the loop circumference - if this argument is correct!

Using the formula for L given in post 16, I obtain:

[itex]\frac{F}{s} = \frac{\mu_0 I^2}{2 \pi r} (ln\frac{8r}{a} - \frac{3}{4})[/itex].
Jan8-13, 09:04 PM
P: 13
Hey thank you again Philip, thinking about the force that way is insightful.

I wanted to let everyone know that I added your integral from #11 to the javascript calculator:

You can test it by clicking the "these values" link. It appears to be very numerically accurate, although I haven't taken the derivative yet for proof. I'm working on an idea for a test using wire thickness.

Zack Morris
Jan8-13, 10:39 PM
P: 13
After mulling over this problem for another couple of days and not getting anywhere, I decided to go back to basics and look at the magnetic force between moving charges:

[itex]\bar{F} _{mag} = \frac{\mu_0}{4\pi} \frac{q _{1} q _{2}}{r^2} \bar{v _{1}} \times (\bar{v _{2}} \times \hat{r})[/itex]

Where [itex]\bar{F} _{mag}[/itex] is the force vector, [itex]v _{1}[/itex] and [itex]v _{2}[/itex] are the velocity vectors of the charges respectively, and [itex]\hat{r}[/itex] is the unit vector from [itex]q _{1}[/itex] to [itex]q _{2}[/itex]

I'm going to list the contributions [itex]F _{12}[/itex] and [itex]F _{21}[/itex] from Tipler Physics 25-3a and 25-4b because I'm having trouble finding them on the web:

[itex]F _{12} = q _{2} v _{2} \times B _{1} = q _{2} v _{2} \times (\frac{\mu_0}{4\pi} \frac{q _{1} v _{1} \times \hat{r} _{12}}{r^2 _{12}})[/itex]

[itex]F _{21} = q _{1} v _{1} \times B _{2} = q _{1} v _{1} \times (\frac{\mu_0}{4\pi} \frac{q _{2} v _{2} \times \hat{r} _{21}}{r^2 _{21}})[/itex]

I'd like to integrate these inside a rectangular cross section ring and also a torus for a given charge velocity, which can be derived from current:

[itex]I = n q v _{d} A[/itex]

So basically, if we assume an even charge distribution, we could calculate the sum of the forces from the other charges on a single charge and then sum those to get the total force.

I might just do this numerically with a finite number of point charges evenly distributed in the ring for now as a check for Philip's #19 formula.

The $64,000 question for me is whether the outward force depends on the cross sectional area A of the ring. I would also like to know if the shape of the area around the centroid matters (my guess is that it won't). Hopefully we can get a thin wire approximation out of this for ring radius >> thickness.

Zack Morris
Andrew Mason
Jan9-13, 07:07 AM
Sci Advisor
HW Helper
P: 6,683
Quote Quote by zmorris View Post

[itex]F _{21} = q _{1} v _{1} \times B _{2} = q _{1} v _{1} \times (\frac{\mu_0}{4\pi} \frac{q _{2} v _{2} \times \hat{r} _{21}}{r^2 _{21}})[/itex]
The Lorentz force for current: F = ILB = B∫jLdA = B∫jdV where j is the current density in the conductor, L is the length of conductor, A the cross-sectional area, dV is the volume differential and B is the magnetic field external to L (ie. it does not include the field created by the current flow in L). To use this, one would to know the current density as a function of volume. I don't think you have to determine the drift velocity of the electrons.

[itex]I = n q v _{d} A[/itex]

So basically, if we assume an even charge distribution, we could calculate the sum of the forces from the other charges on a single charge and then sum those to get the total force.
You are making it more difficult than it is. The difficult part is determining B, not I or the Lorentz force. Just use dF = IBdL or dF/dL = IB where B is determined by the Biot-Savart formula.


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