How to prove that the net force on an irregular current carrying loop in a uniform magnetic field is zero?

I can prove it for regular shaped body like circular loops or rectangular loops.But how to prove it for any arbitrary shaped closed loop? Any suggestions will be appreciated.

(To be clear, this is only referring to linear force, not torque.)

  • $\begingroup$ By how much does the field energy change when you displace the loop? $\endgroup$ – CuriousOne Sep 21 '15 at 19:18
  • $\begingroup$ Potential energy is negative of dot product of magnetic moment and magnetic field vector,as far as I know @CuriousOne $\endgroup$ – user74370 Sep 21 '15 at 19:19
  • $\begingroup$ So how does it depend on a displacement of the loop when the field is homogeneous? What does that tell you about the force? $\endgroup$ – CuriousOne Sep 21 '15 at 19:22
  • $\begingroup$ Force can be given as negative gradient of potential energy I guess.But I'm not getting it :-( @CuriousOne $\endgroup$ – user74370 Sep 21 '15 at 19:26
  • $\begingroup$ How does a homogeneous field change when you move around in it? Is it the same everywhere or does it depend on the position? I can't make it any more simple than that. $\endgroup$ – CuriousOne Sep 21 '15 at 19:33

How to do it the hard way

The Lorentz-force on a curve $C$ carrying a uniform current I is given by the line integral: $$\vec F = I \int_C d\vec \ell \times \vec B$$ For a constant $\vec B$-field over a closed loop, this reduces to: $$ \vec F = I \left[\oint_C d\vec r \right]\times \vec B = I \vec 0 \times \vec B = \vec 0.$$ The key factor here is that the line integral of $1$ over a closed loop is $\vec 0$ no matter the shape of the loop. It is easiest to see this component-by-component; the $x$-component for example is $\oint_C dx = \Delta x,$ but any net displacement in the positive direction needs to eventually be compensated by net displacement in the negative direction, to bring it back to where it started.

Magnetic force on a current-carrying wire

Why the heck is that first expression true? You just have to be really meticulous about what you mean, basically. A general current loop is a parametric curve $C = \{~\vec r(s) \text{ for all } s \text { in } (0, S)~\}.$ Between two points $s, s+ds$ you'll see a tangent vector $d\vec r = \vec r'(s) ~ds.$ There is probably a linear charge density for the moving charges $\lambda(s)$ over the loop, and they move at a speed $v(s)$ such that $\lambda(s)~v(s) = I(s)$ is the current through the wire: we usually want this to be a constant, so charge isn't "building up" at any point, otherwise that creates $\vec E$ fields which oppose the $\vec B$-field forces involved.

We don't really have a time coordinate as we're only looking at one moment in time, but let me take a typical charge flow over parameter-change $ds$ as happening over a real-time-change $dt$. Then, identifying $v(s)$ as $\left|\frac {d\vec r}{dt}\right| = \left|\vec r'(s)\right| \cdot \left|\frac{ds}{dt}\right|$ we get $\frac{ds}{dt} = I/\big[\lambda(s)~ |\vec r'(s)|\big].$ So that gives an explicit notion for $dt$.

The net Lorentz force due to the magnetic field is of course $$\vec F = \oint_C dq~\vec v\times\vec B, $$and we identify $dq = \lambda(s) ~ |\vec r'(s)|~ ds$ and $\vec v = \frac {d\vec r}{dt} = \vec r'(s) ~\frac{ds}{dt}$ to turn this into: $$\vec F = \int_0^S ds~I(s) ~ \vec r'(s) \times\vec B(s) = I \oint_C d\vec r\times \vec B.$$ So you just get a straightforward line integral.

Another simple way to see this is as the curvy generalization of the well-known result for a length $L$ of wire oriented in direction $\hat n$ with current $I$ going in the $\hat n$ direction: then the force on the wire is $\vec F = L~I~\hat n\times\vec B.$ (Some people write $\vec I = \hat n ~I.$)


Simply $$F =BIL$$ IN A CLOSED LOOP the $\vec{L}=0$ so .F=0

  • 1
    $\begingroup$ $$\vec F = I \int_C d\vec \ell \times \vec B$$ $\endgroup$ – Unique Jul 8 at 4:26

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