How to Find Magnetic Field Components of a Circular Loop?

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field components (Br and Bθ) and the vector potential (A) of a circular loop with radius R, centered at the origin, carrying a counter-clockwise current. For point Q=(a,0,a) where a>>R, the azimuthal symmetry allows simplification, leading to the conclusion that Bφ=0, Br is derived from Aφ, and Bθ is calculated using the appropriate derivatives. For point Q=(ha,0,0) with h<1, the vector potential A can be expressed as a multipole expansion, confirming the use of associated Legendre polynomials in the series.

PREREQUISITES
  • Understanding of magnetic field theory and vector calculus
  • Familiarity with the Biot-Savart law and its applications
  • Knowledge of multipole expansions in electromagnetic theory
  • Proficiency in using associated Legendre polynomials
NEXT STEPS
  • Study the Biot-Savart law in detail for various geometries
  • Learn about multipole expansions in electromagnetic fields
  • Explore the derivation and applications of associated Legendre polynomials
  • Investigate the implications of azimuthal symmetry in magnetic field calculations
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as engineers working with magnetic field applications and simulations.

shinobi20
Messages
277
Reaction score
20

Homework Statement


A circular loop of radius R is on the xy plane and the center is at the origin, the current is flowing in a counter-clockwise manner. a) Let Q=(a,0,a) be a point such that a>>R. Find Br and Bθ at Q. b) Let Q=(ha,0,0) be a point on the x-axis such that h<1. Find the vector potential A at Q as a power series of h.

Homework Equations


A(r) = k∫ (J(r') dτ') / |r-r'| = kI ∫ dr' / |r-r'| where k is μo/4π and I is the current

The Attempt at a Solution


a) From azimuthal symmetry, we can restrict the situation to points r on the xz plane.
dr'=(dx', dy', 0)=(-Rsinφ', Rcosφ', 0)dφ'. Since the only non vanishing component of A is Aφ

Aφ(r) = kI ∫ (Rcosφ' dφ') / |r-r'| from 0 to 2π

Bφ = 0
Br = - 1/r ∂/∂cosθ (Aφsinθ)
Bθ = - 1/r ∂/∂r (rAφ)

Is this correct?

For part b) I don't know if it is a multipole expansion or somethin' else... Any help?
 
Physics news on Phys.org
I think for part b) it can be expressed as a multipole expansion, Aφ = kI ∑l=0 ( ha< Pl(0) Pl(cosθ) )/( l(l+1) ). Is this correct?
 

Attachments

  • Snapshot.jpg
    Snapshot.jpg
    10.9 KB · Views: 468

Similar threads

Replies
12
Views
2K
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
27K
  • · Replies 5 ·
Replies
5
Views
16K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K