B-Fields do no work?

  • #1
ChanceLiterature
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This statement has always confused me. Now my confusion is coming home to roast while we cover EMF.

If we consider work mathematically as ∫f.dl and require integral to be path independent then of course the B-field does no work.

However, it seems like there is a deeper meaning to B-fields do no work. Is there?

Tied into this is faradays law. Faraday's Law can be "derived" from emf (it's in quotes because I understand faradays is first principles). In this derivation emf force is stated to be closed integral ∫E.dl. I do not understand why this why emf or this induced force around the loop is necessarily due to an E-field as opposed to to B-field.
 
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  • #2
anuttarasammyak
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For some clarification I think you are saying that B-field do no work to charges. B-field do work to magnetic dipoles as we see two magnets attract and collide.
In this derivation emf force is stated to be closed integral ∫E.dl.
it equals to
[tex]-\int_A\frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}[/tex]
using B.
 
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  • #3
ChanceLiterature
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For some clarification I think you are saying that B-field do no work to charges. B-field do work to magnetic dipoles as we see two magnets attract and collide.

it equals to
[tex]-\int_A\frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}[/tex]
using B.
Yes, I'm trying to highlight two points of confusion. You're right I was not clear enough. I hope this helps.

1) I am asking why B-fields do no work on a point charge when the path integral of a charge on a B-field is non-zero? Did we define work as path integral only over conservative fields? Is this why we say B-fields do no work?

2) I am not sure why the closed line integral of B is necessarily zero (if I still had my copy of Boas' Mathematics in the physical sciences, I might know why/if this holds). Additionally, if it is not necessarily zero, I am not sure why the resulting the emf is defined as closed E.dl.

My confusion with Faraday's law in the original question derives from not understanding emf I suspect. Thus, I think ironing out point of confusion 2 will help
 
  • #4
anuttarasammyak
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A familiar reply to 1) is to see the formula of Lorentz force the part originated from B of which is
[tex]q \mathbf{v} \times \mathbf{B}[/tex]
and is always perpendicular to velocity of charge q point, ##\mathbf{v}##, thus no work done on the charge.
 
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  • #5
ChanceLiterature
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Thanks. I realized 1 was very dumb after remembering what conservative and close integral meant.

Additionally, even I'm not sure what I'm asking or saying for 2
 
  • #6
ChanceLiterature
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Thanks for the help
 
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  • #7
anuttarasammyak
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2) I am not sure why the closed line integral of B is necessarily zero
Zero we see as for B is
[tex]\nabla \cdot \mathbf{B} =0[/tex]
and thus
[tex]\int_S \mathbf{B} \cdot d\mathbf{S}=0[/tex]
for a closed surface S. Not line integral.
 

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