Force Due to Magnetic Field on a Charge Carrying Wire

Click For Summary
SUMMARY

The discussion centers on calculating the force on a current-carrying wire in a non-uniform magnetic field using the equation F = iL × B. The magnetic field is defined as B = 3i + 8.0x²j, indicating that it varies with position along the wire. Integration is necessary to compute the force between x=1 and x=3 due to the non-uniformity of the magnetic field. The standard formula F = iL × B applies only when the magnetic field is constant.

PREREQUISITES
  • Understanding of vector calculus, specifically integration
  • Familiarity with the Lorentz force law
  • Knowledge of magnetic fields and their representations
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the application of the Lorentz force law in varying magnetic fields
  • Learn about vector integration techniques in physics
  • Explore examples of non-uniform magnetic fields and their effects on current-carrying conductors
  • Investigate the implications of magnetic field uniformity on force calculations
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and professionals involved in electrical engineering and applied physics.

wk1989
Messages
32
Reaction score
1
Hi, I have a bit of a problem understanding one of the solutions for my assignment.

Homework Statement


Normally we use the equation
F = iL X B
to find the force of a magnetic field on a wire with a current.

One of the questions asks to find the force on a section of wire between x=3 and x= 1, the wire's current is in the the negative x direction. The magnetic field is something like B=3i+8.0x^2

They used integration to solve the problem, integrating the change of force from 1 to 3. I'm wondering why this is used? Is it because the magnetic field is not uniform (it's in terms of x)? If the magnetic field is uniform, could we just have done F = iL x B with 2 being the L?

Thanks in advance!
 
Physics news on Phys.org
Is the magnetic field supposed to be B = 3i + 8.0x^2j?

If that's the case, then, yes, the field is not uniform over the section x = 3 to x = 1, so an integration is necessary over the length of the wire.
 
You can only use [tex]\vec{F} = I (\vec{L} \times \vec{B})[/tex] when the magnetic field is constant. Since here, your magnetic field is different over a particular area, you have to use integration.
 

Similar threads

Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 6 ·
Replies
6
Views
810
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
2K