Magnetic field generated by current in semicircular loop at a point on axis


by SOMEBODYCOOL
Tags: biot-savart, magnetic field, magnetic fields, semi-circle
SOMEBODYCOOL
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#1
Mar13-10, 10:11 PM
P: 5
1. The problem statement, all variables and given/known data
Determine the magnetic field strength and direction at a point 'z' on the axis of the centre of a semi-circular current loop of radius R.


2. Relevant equations
Biot Savart Formula
[tex]d\vec{B}=\frac{\mu_{0}Id\vec{r}\times\hat{e}}{4\pi|\vec{R}-\vec{r}|^{2}}[/tex]

e being the unit vector from r to R
3. The attempt at a solution
A much simpler problem is a full current loop, because one component of the magnetic field cancels out. For this problem, you'd have to deal with the half-circle arc and the straight line base separately. I was also wondering whether its easier to calculate the z and x components of B separately as well... One component is straightforward enough... I just really don't understand where to start.
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gabbagabbahey
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#2
Mar14-10, 12:15 AM
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This should be a pretty straightforward application of the Biot-Savart Law. Start by finding expressions for [itex]\textbf{r}[/itex], the position vector for a general point on the semi-circular arc, and [itex]\textbf{R}[/itex] the position vector for a general point on the [itex]z[/itex]-axis....what do you get for those?...What does that make [itex]\hat{\mathbf{e}}[/itex]? What is [itex]d\textbf{r}[/itex] for a semi-circualr arc?

To makethings easier, you will want to use cylindrical coordinates.
SOMEBODYCOOL
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#3
Mar14-10, 03:23 PM
P: 5
So, the parametric representation of a point on the semi-circle would be (0, bcos(t), bsin(t)) where b is the radius of the semi-circle.
The vector R is just [d, 0, 0] where d is the distance on the axis of the point
and then the e is the unit vector from R-r
But what's dr? And where does the switch to cylindrical coord come in?

SOMEBODYCOOL
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#4
Mar14-10, 04:34 PM
P: 5

Magnetic field generated by current in semicircular loop at a point on axis


I think I got it. Thanks
gabbagabbahey
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#5
Mar14-10, 11:19 PM
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Quote Quote by SOMEBODYCOOL View Post
I think I got it. Thanks
If you'd like to post your result, we''ll be able to check it for you.


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