Question

I'm taking another, unsolved, problem from PhysicsForums that I came across (and am curious about) and am posting it here. Anybody familiar with the Bios-Savart Law?

Answer 1

"An electric current I flows in a straight conductor of length L. Use the law of
Biot-Savart to find the magnetic field at a point lying on an axis going through the
centre of the conductor and perpendicular to the conductor. "

Answer 2

yes, you need to set up an integral !

Answer 3

^ http://sethdcohen.com/wp-content/uploads/2014/03/Captain-Obvious.jpg

Thank you for your untold wisdom, oh great one. I would be unfathomably lost without you.

Answer 4

lol. nice one!

Answer 5

Lol..

well.. sometimes people don't realize that they need to do an integral..

my bad then :P.. where are u getting stuck?!

Answer 6

Besides if that was obvious..

next time don't leave question like that.. show how much work you have done.. and mention specifically where u need help :P :P!

Answer 7

You're totally right, heh, I'm just messing with you. I'm about to head to work, but as soon as I get back from it, I'll respond. Thank you nonetheless.

Answer 8

What I'm confused with is how to express the integral solely in terms of a variable necessary to integrate with respect to, across the length of the conductor. How should I set up dL?

http://i.imgur.com/XacrD5D.png

Answer 9

dl is the length of current element.

In your expression dl is along y, so you shouldn't use y as the variable.. you should call it as l

so then the expression becomes

\[\vec B = \frac{\mu_0i}{4 \pi} \int\limits_{-L/2}^{L/2}\frac{\vec dl \times \vec r}{(l^2+r^2)^{3/2}}\]

Answer 10

Alright, so how do I move from there?I have a variable of integration inside of a cross product.

Answer 11

(?)

Answer 12

Get rid of that cross product :D.


look at the direction. Resolve it. And then think which component matters?

One component will cancel out as you integrate (which you can neglect)
and another component will add up (so you can just integrate and forget about vectors)

Answer 13

Oh, alright, I think I get it, since l and r are always perpendicular, you can the cross product of the vectors is just their magnitudes multiplied together, yes?

Answer 14

It is generally easier to use an angular variable.


Integrate for θ from -α to +α

Answer 15

@Mashy Quote: "look at the direction. Resolve it. And then think which component matters? "

I do not understand this. All elementary vectors are inside the plane, so no resolving is necessary.

Answer 16

In retrospect, I'm still not sure how to evaluate this integral at all, because what's in the numerator is a cross product, not the magnitude of a cross product, anyways.

Answer 17

You must evaluate the cross product as dl.r.sin (angle)

Answer 18

Yes, but what's in the numerator is a vector, not the magnitude of a vector, so it isn't appropriate to evaluate it as such, is it?

Answer 19

If you consider the contribution to the magnetic field from the current element Idl, you will find it is perpendicular to the plane defined by the wire and the axis.
The same is true for ALL the current elements along the wire, each contributing to the field B in that same direction perpendicular to the plane.

In general if you have to integrate something with a vector integrand, you can consider components of the vector, in an appropriate coordinate system, and integrate the expression for each component - so really a vector integral is 3 scalar integrals.

In the problem being considered here, you simply need to integrate to find the total component of the field perpendicular to the plane of the wire and the axis, because the field only has that component.

Answer 20

Just clarified the labelling on the diagram