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

  1. Dec 30, 2012 #1
    Hi, I'm trying to calculate the outward force on a loop of wire carrying a current, radially from the center to the perimeter. I found the formula for the force between two parallel wires:


    F/L = u0*I1*I2/(2*pi*r)

    And the general formula here:


    It's easiest to see the formula in the link because it contains integrals.

    The problem is, I've been out of college too long and my brain is suffering from petrification. I've always had a hard time visualizing line integrals. My guess is that when parallel wires are bent into a ring, each infinitesimally small section of wire will feel an increasing force as the opposite section of wire is bent closer to it. So probably the 1/2pi term will fall away when the wires are formed into a circle. I'd like to work the integral to know for sure, but can't remember how to do it.

    A simpler question related to this is: when we find the force per length on one wire with respect to the rest of the wire, what is the radial force outward with respect to the origin? I think it's half as much because the force on each radius added together would total the force across the ring.

    So my best guestimate is that the radial force per length anywhere on the ring is:

    F/L = u0*I/2

    But it seems that the "elegant" solution would be:

    F/L = u0*I

    Can anyone do better?

    Thanks for your help,

    Zack Morris
  2. jcsd
  3. Dec 31, 2012 #2

    Andrew Mason

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    Since the net force is radial, by symmetry all the force components other than the radial components cancel each other out and you end up with just the radial components. So try integrating the radial force on an element dL = rdθ around the loop using dF = IBdL

  4. Dec 31, 2012 #3
    That's a good approach and has me thinking in some new directions. Unfortunately I don't know how to calculate the B field strength at the opposite side, because I only have the formula for a straight wire. So I don't have anything to plug in to IBdL.

    Another way to think of this is, imagine we had an infinite wire bent into a parabola with the focus at a point on an opposite wire. The field would have to be stronger there than for the straight wire case. I just don't know by how much. Like maybe I could calculate the strengths of two half circles that way or something and add them.

    I'm running into some other problems too, where the self-field strength is infinity at the center of the wire one one hand, but zero on the other, since the current runs parallel to the wire and the cross product becomes zero there.

    After two days of thinking about this, I'm just not getting anywhere. I may try a numerical solution using the formula for the B field of a wire segment:

    B = (uo/2pi)*(1/R)*(sin A1 + sin A2)

    Where A1 and A2 are the angles from the ends of the wire segment to the point, and R is the perpendicular distance from the wire segment to the point.

    Unfortunately no matter how many times I subdivide the circle into segments, the segments near the point on the circle have R of 0. My guess is that since the sin of a small angle approximates the angle, the (sin A1 + sin A2)/R term might cancel out to some number (possibly 0, 1 or 2?) for small A.

    If I was better at line integrals, or even polar integrals for that matter, I could plug in that formula, calculating A1 and A2 for some point (say x = 1 on a unit circle on the xy plane) and find the total B field there.

    Once I have the B field strength at any point on the circle, I can plug in the current to get the force per unit length there.

    I feel like I am missing something very basic here though, say like the B field being constant across the area of the circle. If that were the case, then I could use that value to find the force on a point on the circle. If I even knew the shape of the B field over the face, it would help.

    I haven't been able to find a formula providing B for arbitrary r inside the circle. I've been sifting through my old college books with no luck too. Very weird.

    Zack Morris
  5. Jan 1, 2013 #4

    Andrew Mason

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    What you are trying to find is the force on a section of wire dL = rdθ due to the magnetic field of its counterpart dL diametrically opposite (ie. a distance 2r away). That magnetic field, B, is determined from Ampere's law using: [itex]\int B\cdot dL = \mu_0I[/itex] which, for a circle of radius 2r means that [itex]B = \mu_0I/2\pi(2r) = \mu_0I/4\pi r[/itex]

    So [itex]dF = IBdL = \mu_0I^2 rd\theta/4\pi r = \mu_0I^2 d\theta/4\pi[/itex]. Just calculate the integral of dF around the circle to find the total outward force.

    The key is recognizing that the forces are radial and you can neglect the forces from sections that are not opposite because they all cancel each other out. At least I think that is right - that is the way it looks to me.

  6. Jan 2, 2013 #5

    Philip Wood

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    AM: I hesitate to differ, but I don't think that all contributions except from the diametrically opposite portion of the circle do cancel. Take, for example the portions, X and Y (say) [itex]\pm\frac{\pi}{2}[/itex] round from the from the portion Z we're considering the force on. Applying the right hand grip/corkscrew rule to X and Y show they produce fields in the same direction at Z.

    Do say if I've got this wrong, or am missing something.

    In my opinion this problem is far from trivial. It's akin to the problem of working out the inductance of a loop. The field is certainly not uniform across the enclosed area (as it is for a solenoid).
  7. Jan 3, 2013 #6

    Andrew Mason

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    If they didn't cancel, then there would be a net force tangential to the loop. By symmetry, the outward force must be entirely radial.

    I am not saying that the total magnetic field at the loop is μ0I/2r. That is just the field from the opposite side. When using the Lorentz force dF = IdLxB the B term does not include the magnetic field created by the current I in the conductor experiencing the force being measured.

  8. Jan 3, 2013 #7

    Philip Wood

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    For what it's worth (which isn't very much) the fields at Z due to X and Y (see post 5) are both at right angles to the plane of the loop, and re-inforce. This does make the force due to X and Y on Z, radial.
  9. Jan 3, 2013 #8

    Andrew Mason

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    Thanks Philip. You are quite correct. I was trying to simplify it but it looks like there is no avoiding a messy Biot-Savart integral here.

  10. Jan 3, 2013 #9
    Hi sorry for my late reply. I bit the bullet and wrote a script to calculate a numerical estimate of the B field at any x,y,z coordinate near a ring of current. It also displays the ideal fields at the center and along the z axis for reference:


    My best guess is that the B field is the same strength as at the center, so 2 pi times stronger than the parallel wire case. If my math is right, the force per length should also be 2 pi times stronger, which is more than I expected. The total outward force would be force per length times circumference.

    My rational for this is: try plugging in x=1 to find the strength of the field on the ring. You can also experiment with turning off the 8 point average and using start and end angles of say 0.01 and 2 pi - 0.01. The estimate hovers very close to the strength of the B field at center.

    I'm more of a programmer than mathemagician so if someone spots any mistakes, please let me know and post the fix.

    If someone knows how to calculate the exact integral for this, they get super bonus points!

    Zack Morris
  11. Jan 3, 2013 #10

    Andrew Mason

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    To calculate the outward force on the loop you must first calculate the magnetic field at a point on the loop produced by an infinitesimal section dL using the Biot-Savart law and integrate. The Biot-Savart law is easy enough to state:

    [tex]dB= \frac{\mu_0 I d\vec{L}\times\hat{s}}{4\pi s^2 } [/tex]

    where s is the magnitude of the displacement of dL from the point and [itex]\hat{s}[/itex] is the unit vector in the direction of that displacement.

    Once you find B at the circumference (ie the loop), which will be the same at each point on the loop, the force can be found from the Lorentz force law for current: F = ILxB = ILB since B is perpendicular to the plane of the loop.

    It is easy to calculate B at the centre of the loop because of the symmetry which makes the integral very simple. But it is not so simple to do the calculation of the field at a point, p, on the circumference. You have to express [itex]d\vec{L} \times \hat{s}[/itex] as a function of angle and integrate. But since s is not the radius of the loop but is the distance between p and dL, it and dL x s keep changing.

    [itex]s = \sqrt{(r+rcos\theta)^2 + (rsin\theta)^2)}[/itex] where θ is the measured from the diameter through the point opposite p and r is the radius of the loop.

    dL = rdθ

    and [itex]d\vec{L}\times\hat{s} = dL\sin\alpha[/itex] where [itex]\alpha[/itex] is the angle between dL and s.

    Once you work out the relationship between [itex]\alpha \text{ and } \theta[/itex] you then have to integrate that horrendous thing from θ = 0 to 2∏. Good luck with that!

  12. Jan 4, 2013 #11

    Philip Wood

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    Well, fools rush in where angels fear to tread. I found both setting up the integral and integrating it (courtesy of Wolfram) ominously straightforward. Unfortunately it gives an infinite value for the field at at a point on the loop, due to the whole loop. We know this not to be the case. I fear that we must consider the finite thickness of the wire - and that is a path down which I shall not venture...

    [Note: in the thumbnail I should have called the current element Ia d[itex]\phi[/itex].
    I should also have stated that the fields at P due to all the elements are in the same direction, namely out of the page if the current is anticlockwise.]

    Attached Files:

    Last edited: Jan 4, 2013
  13. Jan 5, 2013 #12
    Ya I came to the same conclusion, that the field depends on the thickness of the wire, because the strength inside is proportional to radial distance from the center over radius of the wire squared:


    Again, everything breaks down for a 0 thickness wire. I think the self field will be 0 at the center of any wire, regardless of thickness though.

    However, we know that there is definitely a force from the rest of the ring because we can calculate it for parallel wires, and we know that the force MUST be greater than or equal to the parallel case because a bent loop will have some interaction with the opposite side (like the focus of a parabola), and the bent neighbors of an infinitesimal length will have little to no interaction because they are nearly parallel to the segment. I'm confident that force per length will be somewhere between 1 and 2 pi times stronger than for the parallel case.

    We might have some luck integrating the force on a moving charge though:


    Because we could calculate the interaction for every charge against the others, regardless of the geometry of the wire.

    I realize I'm getting kind of obsessed over this, but I have looked for days now in both my physics books and the web and have not found any satisfactory answers. If this was a couple of centuries ago, we would just measure the outward force on a loop and see if there was any dependence on wire thickness. I'm sure we'd find that there isn't though, because wire thickness is not present in the parallel wire force formula.

    I'm concerned because if we had this basic outward force formula, we could use it to derive the other formulas. Without it, we're forced to integrate solutions over and over again, and I think it's a distraction from something fundamental. I'm just picturing how messy it is as the dimensions shrink and we picture things like electron clouds. This is the first time I've really run into major problems with singularities, and I could easily see how someone would stray into feel-good solutions like string theories to get around them (basically sampling at multiple points to get around infinities). If we can't even solve this, then something has been overlooked in the formulas (or at the very least, my own attempt at an education). I wish I had the time and money to research this or at least simulate it more in a computer.

    On that note, I'm thinking I might set up a line integral page so you could enter in piecewise smooth segments and the formula you want to analyze and it would calculate approximations for you so you could check your integrals.

    I'm going to list Philip's solution here to build from:

    B = (u0*I/16*pi*r)*2*[ln(tan(phi/4))] from phiStart to phiEnd

    So for a complete circle, it's from 0 to 2pi. But tan of 2pi/4 is infinity.

    Thanks for everyone's help, I still think there is hope for this problem if we could just see it,

    Zack Morris
  14. Jan 5, 2013 #13

    Philip Wood

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    For an infinite parallel wire, Ampère's circuital law (+ symmetry) shows that the field outside the wire is the same as if all the current flowed along the axis of the wire. I suspect that the motor effect force on the other wire is as if the current flowed along its axis, but I'm not sure on this point. Any departure from the force calculated assuming axial flow would certainly be very small if the wires are much further apart than their diameters.

    The problem with the loop integration arises near the limits ([itex]\phi = 0, \phi = 2\pi).[/itex] The r2 term on the bottom drives the integrand towards infinity more powerfully than the sin[itex]\frac{\phi}{2}[/itex] on top drives it towards zero.
  15. Jan 5, 2013 #14

    Andrew Mason

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    You have done a nice job there.

    I occurs to me that the Biot Savart law as I have written it is not really applicable when you get close to P (ie. r → 0) because the magnetic field reaches a maximum just outside the wire (B = μ0I/∏d where d is the diameter of the wire) and when you go to smaller distances you have to take into account current density in the wire. When you get inside the wire there is no enclosed current so the B field is 0. So when the angle in your integral is such that line from dl to P passes entirely through the wire, the field contribution ≈ 0. You could work out when that would occur (call it angle δ) and just set the limits of integration from 0+δ to 2∏-δ. For starters, just let δ=2∏/360 and try running that through your calculators.

    Last edited: Jan 5, 2013
  16. Jan 6, 2013 #15


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    This reminds me of my first attempt at calculating the inductance of a wire loop, and assuming an infinitely thin wire. Magnetic flux diverges due to the 1/r dependence of B near the wire. It surprised me because calculating capacitance is so straightforward, in comparison.

    Calculating the magnetic field of a wire loop with finite-diameter wire is on my bucket list.
  17. Jan 6, 2013 #16

    Philip Wood

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    Redbelly: I'm just wondering whether it's possible to relate the outward force per unit arc length (or, equivalently, the loop tension) to the loop inductance, by a virtual work type of argument. Wouldn't solve either problem, but would show them to be essentially one and the same - if my hunch is right.

    Incidentally, I've found the solution to the loop inductance problem in J H Jeans (Yes, he of Rayleigh-Jeans fame) The Mathematical Theory of Electricity and Magnetism (1927). It's not particularly easy, and I haven't digested it yet. It proceeds via mutual inductance, and it does, of course, take account of the wire thickness, which indeed appears in the final formula.

    [itex]L = \mu_0 r(ln \frac{8r}{a} - \frac{7}{4})[/itex]

    r is the loop radius and a is the wire radius. The current is assumed to be uniform across the wire's cross-section.

    Hope this hasn't spoilt the fun for anyone.
    Last edited: Jan 6, 2013
  18. Jan 6, 2013 #17
    Hey thanks Philip, solving for inductance instead was a brilliant insight. Here is a similar formula in "On the Self-Induction of Electric Currents in a Thin Anchor-Ring" by Rayleigh 1912 I found after you mentioned it:


    The formula is:

    [itex]L = 4\pi r(ln \frac{8r}{a} - \frac{7}{4} + \frac{a^2}{8r^2}(ln \frac{8r}{a} + \frac{1}{3}))[/itex]

    Where r is the loop radius and a is the wire radius, the same as Jeans, and current is assumed to be uniform across the wire's cross-section also.

    I think that the [itex]4\pi[/itex] term in this represents [itex]\mu_0[/itex] before it was common.

    I'm wondering where the discrepancy of the [itex]+ \frac{a^2}{8r^2}(ln \frac{8r}{a} + \frac{1}{3})[/itex] term comes from. It looks like it approaches 0 as r >> a. I dropped it into google and got 0.0059 for r = 10a and 0.000088 for r = 100a.

    Assuming these formulas are correct, I'm looking for a formula that relates the angular momentum of the electrons in the inductor to centripetal force. Another way to say that is, the energy stored in the inductor is:


    [itex]E = \frac{1}{2}L I^2[/itex]

    So we should be able to find the total outward force from this somehow.

    A poor analogy of this is, imagine we have a disk with some volume of gas inside, and we add energy. What is the outward pressure for a given amount of energy?

    What I'm getting at is, I'm trying to find the relationship between the centripetal force on the electrons in a magnetic field and the electromagnetic force. I'm thinking the electromagnetic force dwarfs the centripetal force because the mass of the electrons is so small, but I'm trying to find the exact formula. I want to know if it's constant, exponential, relativistic, what. I honestly thought this was going to be straightforward but the rabbit hole keeps getting deeper!

    P.S. Thanks Redbelly and Philip for teaching me Latex formatting.

    And thanks everyone,

    Zack Morris
    Last edited: Jan 6, 2013
  19. Jan 6, 2013 #18
    Just to be clear, my goals are:

    1. Find the total outward force on a ring of current.
    2. Relate this to the centripetal force of the electrons in the ring.
    3. Relate this to the electromagnetic force on the electrons.
    4.* A full formula involving the radius or cross sectional area of the wire would be good, but I also want to see what the limit is as it approaches 0.

    Zack Morris
  20. Jan 8, 2013 #19

    Philip Wood

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    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].
    Last edited: Jan 9, 2013
  21. Jan 8, 2013 #20
    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
    Last edited: Jan 8, 2013
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