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, 2015 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, 2015 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, 2015 at 19:22
  • 1
    $\begingroup$ @CuriousOne But segments of current-carrying wire do experience a force when moved in a magnetic field. $\endgroup$
    – rob
    Sep 21, 2015 at 19:48
  • 1
    $\begingroup$ You should start by assuming the wire isn't moving (otherwise it isn't necessarily true). You might also want to assume the wires are very very thin. $\endgroup$
    – Timaeus
    Sep 21, 2015 at 21:41

2 Answers 2


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

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    $\begingroup$ $$\vec F = I \int_C d\vec \ell \times \vec B$$ $\endgroup$ Jul 8, 2019 at 4:26

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