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

Click For Summary
To determine the magnetic field strength and direction at a point on the axis of a semi-circular current loop, the Biot-Savart Law is applied. The problem is simplified by recognizing that the magnetic field components from a full loop cancel out, but the semi-circular arc and the straight line base must be analyzed separately. Using cylindrical coordinates is recommended for easier calculations, with the parametric representation of points on the semi-circle provided. The discussion emphasizes the importance of defining the position vectors and unit vectors correctly for accurate calculations. Participants are encouraged to share their results for verification.
SOMEBODYCOOL
Messages
5
Reaction score
0

Homework Statement


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.

Homework Equations


Biot Savart Formula
d\vec{B}=\frac{\mu_{0}Id\vec{r}\times\hat{e}}{4\pi|\vec{R}-\vec{r}|^{2}}

e being the unit vector from r to R

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.
 
Last edited:
Physics news on Phys.org
This should be a pretty straightforward application of the Biot-Savart Law. Start by finding expressions for \textbf{r}, the position vector for a general point on the semi-circular arc, and \textbf{R} the position vector for a general point on the z-axis...what do you get for those?...What does that make \hat{\mathbf{e}}? What is d\textbf{r} for a semi-circualr arc?

To makethings easier, you will want to use cylindrical coordinates.
 
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?
 
I think I got it. Thanks
 
SOMEBODYCOOL said:
I think I got it. Thanks

If you'd like to post your result, we''ll be able to check it for you.
 

Similar threads

Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
648
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
7K