How Does the Biot-Savart Law Apply to Uneven Currents in a Split Loop?

  • Thread starter Thread starter GDGirl
  • Start date Start date
  • Tags Tags
    Law Loop Split
Click For Summary
SUMMARY

The discussion focuses on applying the Biot-Savart Law to calculate the magnetic field at the center of a circular conducting ring with uneven current distribution. The ring has a radius of 10.7 cm, with I1 = 3.8 A flowing through the top semicircle and I2 = 10 A through the bottom semicircle. The correct approach involves calculating the magnetic fields for each semicircle separately, taking into account the direction of the currents, which affects the signs of the resulting magnetic fields. The final magnetic field at the center is determined by vector addition of the two fields, considering their directions.

PREREQUISITES
  • Understanding of the Biot-Savart Law
  • Knowledge of vector addition in physics
  • Familiarity with magnetic fields generated by current-carrying conductors
  • Basic calculus for evaluating integrals related to circular paths
NEXT STEPS
  • Study the Biot-Savart Law in detail, focusing on its application to circular currents
  • Learn about vector fields and how to perform vector addition
  • Explore the concept of magnetic field direction and the right-hand rule
  • Investigate the effects of varying current distributions on magnetic fields
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electromagnetism and magnetic field calculations.

GDGirl
Messages
49
Reaction score
0

Homework Statement


A circular conducting ring or radius R = 10.7 cm is connected to two exterior straight wires ending at two ends of a diameter (see Figure). The current splits into uneven portions, with I1 = 3.8 A passing through the top semicircle, and I2 = 10 A passing through the lower semicircle. What is B at the center of the ring?
HELP: Apply the Biot-Savart Law to each semicircle. Adding the two resulting B fields, being careful to keep track of their signs.
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/14/02/P28_31.jpg

Homework Equations


B=\mu0I(\pir)/4\pir2 (for each half circle)

The Attempt at a Solution


I used the equation I put above, and found that the B field for the first current was 1.114e-5 (B1=(4\pix10-7)(3.8)/(4\pi)(.10702 (\pi.107)) and the second current gave me 2.935e-5. From there I tried simply adding them, to give me a total of 4.049e-5. This is incorrect. So I figured they might be vectors, and tried using pythagorean on them (\sqrt{(1.114e-5)^2+(2.935e-5)^2}) and got 3.139e-5. this is also incorrect.
I figure I'm doing something wrong, but I'm not sure what. The hint says to be sure to pay attention to the signs of the B fields, but I don't see where I would get something other than a positive sign.
 
Last edited by a moderator:
Physics news on Phys.org
What is the method of finding the direction of the magnetic field due to current carrying conductor?
Since the directions of the current in two semicircles is not the same, the signs of the fields must be different.
 
rl.bhat said:
What is the method of finding the direction of the magnetic field due to current carrying conductor?
Since the directions of the current in two semicircles is not the same, the signs of the fields must be different.

But aren't the directions the same? Both currents are going to the right.
 
In semicircle direction should be either clockwise or counterclockwise.
 
Oh. Okay, I got it now, thanks. :)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 18 ·
Replies
18
Views
5K
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
3K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
298
Replies
1
Views
2K