Circular loop in uniform magnetic field

[Trollfaz]

Consider a circular loop with uniform current flowing around it in a uniform magnetic field.
Does it experience no translational force due to its symmetry

 

[Baluncore]

A circular loop with uniform current flowing around it will generate a magnetic field.
What is the orientation of the loop with regard to the external uniform magnetic field.

 

[Meir Achuz]

The force would be given by $\vec F=\oint I{\vec dr}\times \vec B=0$ if I and ${\vec B}$ are constant.
Please help with LateX.

 

[Orodruin]

The force would be given by $\vec F=\oint I{\vec dr}\times \vec B=0$ if I and ${\vec B}$ are constant.
Please help with LateX.

… and therefore, following an integral theorem and using $\nabla \cdot \vec B = 0$ (and assuming I didn't screw up the index calculus due to baby on one arm ... *),
$$
\vec F = \int_S (\nabla \vec B) \cdot d\vec S
$$
where $S$ is a surface bounded by the loop. For a spatially constant magnetic field, the net force therefore becomes zero (due to the derivatives being zero).

Hence, the result of the net force being zero if the magnetic field is spatially constant is independent of any symmetry of the current carrying loop.

* The conclusion remains the same regardless of if the index calculus is correct or not - there will only be derivatives of $\vec B$, which equal to zero.

 

[vanhees71]

It's easier to clarify using a continuous current density. The cartesian components of the force are given by
$$F_k=\int_{V} \mathrm{d}^3 x \vec{e}_k \cdot (\vec{j} \times \vec{B}) = \int_V \mathrm{d}^3 x \epsilon_{klm} j_l B_m.$$
Here $V$ is a volume which encloses the entire current distribution, i.e., $\vec{j}=0$ outside this volume and along its boundary, $\partial V$.

Now, because for stationary currents $\vec{\nabla} \cdot \vec{j}=\partial_k j_k=0$
$$j_l = \partial_n (x_l j_n)$$
and thus
$$F_k = \int_V \mathrm{d}^3 x \epsilon_{klm} \partial_n (x_l j_n) B_m.$$
Since further $\partial_n B_m=0$ for a homogeneous magnetic field, you have, using Gauss's integral theorem
$$F_k=\int_V \mathrm{d}^3 x \partial_n (\epsilon_{klm} x_l j_n B_m) = \int_{\partial V} \mathrm{d}^2 f_n \epsilon_{klm} x_l j_n B_m=0.$$

 

[Meir Achuz]

It is interesting to see these tricky ways to show that $\oint\vec dr=\vec 0.$
Since the integral is a vector with no possible direction, it must vanish.

 

[vanhees71]

That's a quick but not too obvious argument ;-)).

 

[Orodruin]

It is interesting to see these tricky ways to show that $\oint\vec dr=\vec 0.$
Since the integral is a vector with no possible direction, it must vanish.

Well, yes … but that makes the generalisation to non-constant fields less obvious.

I’d add that the intuitive way to interpret the integral is the total displacement when following a closed loop around once - hence zero.

 

[Meir Achuz]

Less obvious than what?

 

[vanhees71]

That's a quick but not too obvious argument ;-)).

I meant, it's not immediately obvious, what this simple argument has to do with the original problem.

 

[Meir Achuz]

I meant, it's not immediately obvious, what this simple argument has to do with the original problem.

"Does it experience no translational force". I didn't think it was necessary to prove that the integral was zero, but two previous posts had gone to some lengths to prove what was immediately obvious.

 

[Orodruin]

"Does it experience no translational force". I didn't think it was necessary to prove that the integral was zero, but two previous posts had gone to some lengths to prove what was immediately obvious.

Well, that is a half-truth to be honest. The posts you are referring to were both treating the general case of non-constant magnetic field before specializing to the constant field scenario. This is relevant to the OP’s question in the sense of deducing the general conditions under which the total force is zero - constant field being a sufficient condition but gemerally not necessary.