Proving Force on Arbitrary Wire Same as Straight Wire

[stunner5000pt]

The figure shows a wire of arbitrary shape carrying a current i between points a and b. The wire lies in a plane at right angles to a uniform magnetic field B. Prove that the force on the wire is teh same as that on a straight wire carrying a current i directly from a to b. (Hint: Replace the wire by a series of "steps" that are parllel and perpendicular to the straight line jkoining a and b.)
WEll umm
well the force on the stirahgt wire from a to b is simply
F = iL \cross B = iLB \sin( \theta)
for the arbitrary wire
F = \int idL \cross B
but i is constant and B is constant so
F = iB \int dL = iBL whicvh is the same as the stariaght wire. Is this correct?

 

[Physics Monkey]

See my reply to your other post. What is the integral
<br /> \int d\vec{\ell}<br />
along the curvy wire from \vec{a} to \vec{b}? Does the integral depend on path? Why?

 

[stunner5000pt]

The only reason i can think of an integral not depending on path is because a force is conservative. But why is the force due to a magnetic field conservative? Is it because the energy reamins constant in this system?

 

[Physics Monkey]

You are integrating a differential, what does that tell you about the result of the integration? If you are uncomfortable with this argument and want to be a bit more formal, then try using the fundamental theorem of line integrals on the integral
<br /> \int_C \vec{v}\cdot d\vec{\ell} = \vec{v} \cdot \int_C d\vec{\ell}<br />
where \vec{v} is an arbitrary vector and C is a path from \vec{a} to \vec{b}. Hint: is \vec{v} the gradient of something? What does this result tell you about <br /> \int_C d\vec{\ell} \,\, ?<br />

 

[stunner5000pt]

ok the result of this integration v \oint dl = v \int_{a}^{b} dl = v [l]_{a}^{b} = v (b - a)

im not sure what v is the gradient of, however.

 

[Physics Monkey]

A couple of things and you should have your answer.

First, don't forget that \vec{a} and \vec{b} are vectors. I don't know if you're confused about this, or if you just don't bother to notate it, or if you can't get LaTeX to work. I just wanted to emphasize this point either way.

Second, are you sure you can't find a function that \vec{v} is the gradient of? Hint: the partial derivative with respect to x of your unknown function is v_x, a constant.

Third, your very close to your answer, here and in the other thread. All you need to do is convince yourself (or more properly, the grader) that
<br /> \int_C d\vec{\ell} = \vec{b} - \vec{a}<br />
independent of the path C. Once you have this fact, the answers to both your questions are easy to get.