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

In summary, the conversation discusses finding 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 using the Biot-Savart Law. The conversation also mentions using cylindrical coordinates for easier calculations. The position vectors for a general point on the semi-circular arc and the z-axis are found, as well as the unit vector from R to r. Finally, the conversation concludes with a confirmation of understanding and an offer for further assistance.
  • #1
SOMEBODYCOOL
5
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
[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

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
  • #2
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.
 
  • #3
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?
 
  • #4
I think I got it. Thanks
 
  • #5
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.
 

1. What is a magnetic field generated by current in a semicircular loop?

A magnetic field is a region of space where a magnetic force can be detected. In a semicircular loop, the magnetic field is generated by the flow of electric current through the loop.

2. How is the direction of the magnetic field determined in a semicircular loop?

The direction of the magnetic field in a semicircular loop is determined by the right-hand rule. If you point your right thumb in the direction of the current flow, the direction of your curled fingers will indicate the direction of the magnetic field.

3. What factors affect the strength of the magnetic field in a semicircular loop?

The strength of the magnetic field in a semicircular loop is affected by the amount of current flowing through the loop, the distance from the loop to the point where the field is being measured, and the radius of the loop.

4. How does the position of the point on the axis affect the strength of the magnetic field in a semicircular loop?

The strength of the magnetic field in a semicircular loop is inversely proportional to the distance from the loop to the point on the axis where the field is being measured. This means that the closer the point is to the loop, the stronger the magnetic field will be.

5. Can the magnetic field in a semicircular loop be used for practical applications?

Yes, the magnetic field generated by current in a semicircular loop can be used in various practical applications such as electromagnets, motors, and generators. It is also used in medical imaging technology, such as MRI machines, to produce detailed images of the human body.

Similar threads

  • Advanced Physics Homework Help
Replies
13
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
208
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
207
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
12
Views
692
  • Introductory Physics Homework Help
Replies
7
Views
138
Back
Top