Outward force on current loop (Ampere's Force Law)by zmorris Tags: ampere's force law, force, outward, pressure, radial 

#19
Jan813, 04:49 PM

P: 860

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 backemf 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 nonlinear 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]. 



#20
Jan813, 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: http://jsbin.com/itanag/5/edit 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 



#21
Jan813, 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:
http://teacher.nsrl.rochester.edu/ph...chapter30.html [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 253a and 254b 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: http://en.wikipedia.org/wiki/Electri...nt#Drift_speed [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 



#22
Jan913, 07:07 AM

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