Why is force on a circular current carrying wire zero?

Question

I have a very silly doubt, and it's bothering me. Magnetic Force because of a line current is:
$$mathbf{F_{mag}} = Iintmathrm{d}mathbf{ltimes B}$$
However, in a constant magnetic field, for closed circular loops, force is zero as limits of integration goes from $0$ to $0$. (From one refrence point 0 to back at it again)

My doubt is, the limits of integration for a closed circular loop of radius $R$ should go from $0$ to $2pi R$. Then how is $mathrm{d}mathbf{l}$ integrated over the entire loop zero?!

Joe · Accepted Answer

It depends which integral you calculate, the line integral of a scalar field or the line integral of a vector field. In books usually $ds$ means scalar field and $dtextbf{s}$ means vector field. I will give you example for both for a circular loop.
First you need a parametrization of a circle with radius R in 3D, for example:
$$gamma(s)=(Rcos(s), Rsin(s), 0)$$
Scalar field
$$int_{S} ds = int_{0}^{2pi}|dot{gamma}| ds = int_{0}^{2pi}R ds = 2pi R$$
Vector field
Let $vec{B}$ be some constant vector field then:
$$int_{S} vec{B} cdot dtextbf{s} = int_{0}^{2pi}vec{B} cdot dot{gamma} ds = Rvec{B} cdot int_{0}^{2pi}(-sin(s), cos(s), 0) ds = Rvec{B} cdot (0,0,0) = 0$$
Therefore we indeed get that the line integral of a constant vector field over a circle with radius R is zero.

SteelCubes · Answer

I think you have mis-interpreted what is meant by $doverrightarrow l$ and $dl$
What we use in calculating $F_{mag}$ is $I int{doverrightarrow l times overrightarrow B}$ and not $Iint {dl B}$
I too had a tough time figuring out the difference between two once in the beginning. 
Actually $dl$ is a scalar quantity. It simply means an infinitesimally small unit of length. When we integrate $dl$We have, $int dl = Delta l$  where l is the total length of the wire, rope or any such body.
But, $doverrightarrow l =dxhat i + dyhat j + dzhat k$ 
And integrating it would lead to $Delta xhat i + Delta yhat j + Delta zhat k $, which is the difference in the position vector of initial and final position.
For a closed loop
$int _{overrightarrow{r_i}} ^{^overrightarrow{r_f}}doverrightarrow l = overrightarrow{r_f} - overrightarrow{r_i}$
and by what I guess a closed loop is $overrightarrow{r_i}=overrightarrow{r_f}$, which leaves us with
$int _{overrightarrow{r_i}} ^{^overrightarrow{r_f}}doverrightarrow l = overrightarrow{r_f} - overrightarrow{r_i}=0$

Farcher · Answer

For the simple case of a circular loop.

$displaystyle F= int_0^theta B,I, Rdtheta , sintheta$ and try it for $theta = pi$ and $theta = 2pi$.
Note the symmetry of the system in that an element on the left has an equal magnitude force and opposite in direction to an element on the right producing a net zero force.

Why is force on a circular current carrying wire zero?

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