What is the Magnetic Field at the Center of a Current-Carrying Loop Segment?

Click For Summary
The discussion focuses on calculating the magnetic field at point A, the center of a circular arc in a current-carrying loop. The loop carries a current of 2.5 A and consists of a circular arc from 60° to 360°, along with horizontal and vertical sections. The magnetic field contribution from the arc is calculated as 5/6 of a complete loop, leading to a preliminary value of 3.27e^-5. The vertical section's contribution is determined using trigonometric relationships, while the horizontal section does not affect the field at the center. The discussion emphasizes the importance of integrating the contributions from different segments to find the total magnetic field.
jromega3
Messages
25
Reaction score
0

Homework Statement



A loop carries current I = 2.5 A in the x-y plane as shown in the figure above. The loop is made in the shape of a circular arc of radius R = 4 cm from qo = 60 ° to q = 360 ° . The loop is completed by horizontal and vertical sections as shown.
What is BA, the magnitude of the magnetic field at point A, the center of the circular arc?

Homework Equations



For the loop part...
loopc3.gif


The Attempt at a Solution



Well, that's it if it were a complete loop, but this is 300/360, or 5/6 of a loop. So I take that and divide by 1.2 and I get 3.27e^-5.
Not sure about the other ones. My calculus is rusty at best.

It's hard to visualize maybe, but it's basically a circle from the 60 degree above the horizon all the way till the end, with a straight line going down to the horizontal center line and then a line connecting that line to the 0 degree mark on the curve.

As always, any help is appreciated.
 

Attachments

  • loopc3.gif
    loopc3.gif
    1.9 KB · Views: 514
Physics news on Phys.org
Now you have to find the field due to the vertical section of the loop. By trig. you can see the length of the vertical section is 2sqrt3. The horizontal section does not produce any field at the centerIn the relevant equation take the integration of dx*sintheta/r^2 where dx is a small element of the straight vertical conductor, r is the distance of the element dx from the center and theta is the angle between the conductor and r.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K
Replies
12
Views
2K
Replies
7
Views
2K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 10 ·
Replies
10
Views
2K