Is the scalar magnetic Potential the sum of #V_{in}# and ##V_{out}##

Click For Summary
SUMMARY

The scalar magnetic potential, denoted as V_in and V_out, is continuous across the boundary of a magnetic cylinder, meaning that V^1 - V^2 = 0 at the boundary. This implies that V_in - V_out = 0, indicating that the potentials are equal at the boundary point, typically expressed as V_in = V_out at r=a. The discussion emphasizes that while the potentials remain continuous, the perpendicular components of the electric (E) or magnetic (H) fields may not be continuous due to the influence of surface charges.

PREREQUISITES
  • Understanding of scalar magnetic potential
  • Knowledge of boundary conditions in electromagnetic theory
  • Familiarity with electric (E) and magnetic (H) field concepts
  • Basic principles of surface charge distributions
NEXT STEPS
  • Study the implications of boundary conditions in electromagnetic fields
  • Explore the concept of surface charge distributions in electrostatics
  • Learn about the continuity of potentials in electromagnetic theory
  • Investigate the behavior of electric and magnetic fields at boundaries
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetic theory, particularly those focusing on magnetic potentials and boundary conditions in fields.

happyparticle
Messages
490
Reaction score
24
Hi,
I'm wondering if I have an expression for the scalar magnetic potential (V_in) and (V_out) inside and outside a magnetic cylinder and the potential is continue everywhere, which mean ##V^1 - V^2 = 0## at the boundary. Does it means that ##V^1 - V^2 = V_{in} - V_{out} = 0## ?
 
Physics news on Phys.org
With these boundary conditions, it is understood that the potential(s) or field(s) are to be evaluated at some point on the boundary. e.g. when writing ## V_{in} = V_{out} ##, they often leave out at ## r=a ##, etc., but that is implied.

It is the case with both electrostatic and magnetic (fictitious) surface charges on the boundary, that the potentials are continuous across the surface charge , even though the perpendicular components of the ## E ## or ## H ## fields are not continuous, but are affected by the surface charge. So long as ## E ## or ## H ## remain finite, the potentials will be continuous across the surface charge distribution.
 
  • Like
Likes   Reactions: happyparticle and vanhees71

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 52 ·
2
Replies
52
Views
8K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
538
  • · Replies 2 ·
Replies
2
Views
4K