Solving a University Level EMF Problem: Need Help!

[monty0423]

Hello,

I am trying to solve a problem in university level electromagnetic fields course.
The problem is in the attachment.

My answer is simple, that for the magnetic flux B$\rightarrow$ (vector field B$\rightarrow$),
the Amperian loop will have a current equivalent to 0 because current through Amperian loop i = I - I = 0.
Therefore, B$\rightarrow$ = 0 because μ0*2*$\pi$*ρ*B = i = 0.
However, the solution stated that the force will be applied to the right direction.

Could anybody enlighten me with what is wrong with my reasoning?

 

[aralbrec]

The closed path integral around a loop may be zero but that doesn't mean H is zero everywhere on the path. Using that sort of information profitably depends on taking advantage of symmetry to say H is the same everywhere around the loop.

 

[monty0423]

So basically, my answer would have been correct if the centers of the two conductors in the left were the same, as in, if they were concentric?
How do I apply symmetry in this context?
The B→ is not zero because the superposition, B→ from the two opposite currents don't cancel out with each other due to the current-carrying conductors not being concentric?

 

[aralbrec]

So basically, my answer would have been correct if the centers of the two conductors in the left were the same, as in, if they were concentric?

Yes then you could have argued the component of H *along the path* is the same everywhere on the path and therefore must be zero. You would still have to make another argument (again symmetry) that the perpendicular component of H was also zero. Bits of current density in the cylindrical shell will cause a perpendicular component of B to appear at Ie.

How do I apply symmetry in this context?
The B→ is not zero because the superposition, B→ from the two opposite currents don't cancel out with each other due to the current-carrying conductors not being concentric?

That's right they don't cancel but you have the right idea -- use superposition.

The field due to the wire inside the cylinder is... the field due to the cylindrical conductor is... then add them together. Symmetry will come into play when you have to think about the field due to the cylindrical conductor.