Stokes Theorem: Vector Integral Identity Proof

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Summary:

It is a proof question, but I am struggling to understand what steps I should take next

Main Question or Discussion Point

Hi,

My question pertains to the question in the image attached.
Screen Shot 2020-01-15 at 3.56.25 PM.png


My current method:
Part (a) of the question was to state what Stokes' theorem was, so I am assuming that this part is using Stokes' Theorem in some way, but I fail to see all the steps.

I noted that [tex] \nabla \times \vec F = \nabla \times (\psi \vec B) = \psi \left( \nabla \times \vec B \right) - \vec B \times \nabla \psi [/tex]
Then I thought that [itex] \nabla \times \vec B = \vec 0 [/itex] as [itex] \vec B [/itex] is a constant vector. Simplifying the above expression, we are left with:
[tex] \nabla \times \vec F = - \vec B \times \nabla \psi [/tex] Then I attempted to include the normal vector [itex] \hat n dS [/itex] by 'dotting' both sides with it:
[tex] \left( \nabla \times \vec F \right) \cdot \hat n dS = \left( - \vec B \times \nabla \psi \right) \cdot \hat n dS [/tex]
Then, using Stokes Theorem, integrating both sides over the surface S, and swapping the LHS for the line integral and using the scalar triple product identity for the RHS:
[tex] \left( - \vec B \times \nabla \psi \right) \cdot \hat n dS = \left(\nabla \psi \times \vec B \right) \cdot \hat n dS = \left(\hat n \times \nabla \psi \right) \cdot \vec B dS [/tex]
[tex] \oint \left( \psi \vec B \right) \cdot d\vec r = \iint_S \left(\hat n \times \nabla \psi \right) \cdot \vec B dS [/tex]

What I have there doesn't look too different from the required solution, however I believe that I may have made a grave error along the way. Would anyone be able to provide me with some suggestions/hints on how to proceed?

Thanks in advance
 

Answers and Replies

  • #2
Infrared
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You can re-write your last line as ##B\cdot \left(\oint_C \psi \ dr\right)=B\cdot\left(\iint_S (\hat{n}\times\nabla\psi) dS\right)##. In general if ##v## and ##w## are vectors and ##B\cdot v=B\cdot w## for all vectors ##B##, then ##v## must equal ##w##.
 
  • #3
166
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You can re-write your last line as ##B\cdot \left(\oint_C \psi \ dr\right)=B\cdot\left(\iint_S (\hat{n}\times\nabla\psi) dS\right)##. In general if ##v## and ##w## are vectors and ##B\cdot v=B\cdot w## for all vectors ##B##, then ##v## must equal ##w##.
Many thanks. Yes, that makes sense.
 

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