Force on Current Loop

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Homework Statement


Prove that, in a uniform magnetic field, the net force on a current loop carrying current I is zero.


Homework Equations


Green's theorem, and magnetic force equation


The Attempt at a Solution


My main question is whether I can use Green's theorem to prove this result... it seems rather too simple if I can!

Since F=∫I(dLxB) integrated around the whole loop, let's say the B=(Bx)i+(By)j+(Bz)k and let's parameterize the closed loops as the path L: x(t)i+y(t)j+z(t)k. Then dL= dx i+ dy j + dz k. And dL x B = (Bz*dy - By*dz)i+(Bz*dx-Bx*dz)j+(Bydx-Bxdy)k.

Can I simply break up the integral expression for F into three closed-loop line integrals for Fx, Fy, and Fz respectively? And then can I apply Green's theorem and (using the statement that it's a uniform B-field), I know the derivatives of any component of the magnetic field must be zero... so the line integrals all turn into double integrals over some region, but with 0 as the integrand. Thus the net force in all directions is 0!

Is that allowed? The question called for me to integrate directly...
 

Answers and Replies

  • #2
TSny
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Using the vector identity AxB +CxB = (A+C)xB show ∫(dLxB) = (∫dL)xB

Then interpret ∫dL
 
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Well, I'm not quite sure how to do that proof other than to say that the integral would be an infinite sum and since that identity holds for the sum of two vectors, I'll just extend that to an infinite number of infinitesimal vectors and then apply it to all the little dL's... then it follows easily.

But is the integral of dL (where dL is a vector) equal to 0 or to the length of the loop? I feel like it should represent the length, but then LxB isn't necessarily 0. See, the integral of dL where dL is a piece of arc length would equal the total length of the loop... but if dL is a vector we can't simply integrate it, because it has components and all that!

Also, was my first method valid even if not a direct integral?
 
  • #4
TSny
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The integral is adding up a lot of little displacement vectors. So, it's really representing vector addition of a bunch of vectors that are strung head-to-tail around a closed loop.

That's different than adding up just the magnitudes of the vectors ∫|dL| which would be the total distance around the loop.
 
  • #5
TSny
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1.
My main question is whether I can use Green's theorem to prove this result... it seems rather too simple if I can!

Since F=∫I(dLxB) integrated around the whole loop, let's say the B=(Bx)i+(By)j+(Bz)k and let's parameterize the closed loops as the path L: x(t)i+y(t)j+z(t)k. Then dL= dx i+ dy j + dz k. And dL x B = (Bz*dy - By*dz)i+(Bz*dx-Bx*dz)j+(Bydx-Bxdy)k.

Can I simply break up the integral expression for F into three closed-loop line integrals for Fx, Fy, and Fz respectively? And then can I apply Green's theorem and (using the statement that it's a uniform B-field), I know the derivatives of any component of the magnetic field must be zero... so the line integrals all turn into double integrals over some region, but with 0 as the integrand. Thus the net force in all directions is 0!

Is that allowed? The question called for me to integrate directly...


Yes, you can do that. For example the z-component of the force is proportional to [itex]\oint[/itex](Bydx-Bxdy) around a closed loop in the xy plane. Define a vector V such that Vx = By and Vy = -Bx. Then you can express the integral as [itex]\oint[/itex](Vxdx+Vydy) which is just the line integral of V around the loop. Then Green's theorem may be used to write it in terms of the curl of V over the area of the loop. Since V is a constant vector if B is uniform, you get the desired result.

But note that if you take any one of your terms in F such as [itex]\oint[/itex]Bydx, you can simplify it directly as By[itex]\oint[/itex]dx = 0.
 
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