Vector Potential of Long Solenoid

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SUMMARY

The discussion focuses on understanding the vector potential of a long solenoid as introduced in an introductory Electricity and Magnetism course. Key equations include Ampere's Circuital Law and Stokes' Theorem, which relate the magnetic field B and the vector potential A. The vector potential A is defined as A = (μ₀NI/2)r̂ inside the solenoid and A = (μ₀NI/2)(R²/r)̂ outside the solenoid, demonstrating the relationship between current and magnetic fields. The conclusion emphasizes that the line integral of the vector potential around a closed path equals the magnetic flux through that loop.

PREREQUISITES
  • Understanding of Ampere's Circuital Law
  • Familiarity with Stokes' Theorem
  • Basic knowledge of vector calculus, specifically curl and divergence operators
  • Concept of magnetic flux and its relation to current
NEXT STEPS
  • Study the derivation of Ampere's Circuital Law in detail
  • Learn about Stokes' Theorem and its applications in electromagnetism
  • Explore vector potential in different geometries, such as toroids and loops
  • Investigate the physical significance of magnetic fields generated by currents
USEFUL FOR

This discussion is beneficial for students in introductory physics courses, particularly those studying Electricity and Magnetism, as well as educators seeking to clarify concepts related to vector potentials and magnetic fields.

malindenmoyer
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Let me preface by saying that I am a freshman in an introductory level Electricity and Magnetism course. My professor has assigned this problem, as he briefly introduced the idea of vector potentials, along with curl and divergence operators. I am VERY much lacking in knowledge of any of these concepts and would appreciate guidance. I understand that this site is not supposed to "do the work" for you, but in this case, an explanation would prove very useful.

Problem Statement and Relevant Equations

Recall Ampere's Circuital Law
\oint{B\bullet \partial l}=\mu_0 I_{enc}=\mu_0 \int{J\bullet dA}

But Stokes Theorem states:

\oint{B\bullet \partial l}=\int_{surface}{(\nabla \times B)\bullet dA}

Therefore:

\nabla \times B=\mu_0 J

We can apply similar reasoning to:

\oint{A\bullet \partial l}=\int{(\nabla \times B)\bullet dA}=\int_{surface}{B\bullet dA}=\phi_{enc}

Apply this to a long solenoid.

\oint{A\bullet \partial l}=\phi_{enc} is similar to \oint{B\bullet\partial l}=\mu_0 I_{enc}

Pursue this analogy by drawin the appropriate Amperian loop (radius r) to show:

A=\frac{\mu_0 NI}{2}r\hat{\phi} inside the solenoid, and:

A=\frac{\mu_0 NI}{2}\cdot\frac{R^2}{r}\hat{\phi} outisde the solenoid.

Note that A follows the current direction. Also note that the 1/r dependence gives B=0 outside the coil. Explain and sketch the lines of A in both cases.

Questions Regarding Problem
  1. Is the A in the problem referring to the vector potential?
  2. What is all of this saying?

As I stated previously, I am very new to this and any help is greatly appreciated. Thanks
 
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malindenmoyer said:
Questions Regarding Problem
  1. Is the A in the problem referring to the vector potential?
Yes.
malindenmoyer said:
  1. What is all of this saying?
The upshot is there's an Ampere's law for the vector potential. If you calculate the line integral of the vector potential A around a closed path, it will equal the magnetic flux through the loop.
 

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