# Magnetic Vector Potential prob 5.32 Griffiths

• precise
In summary, the problem at hand involves continuity of the magnetic vector potential and its derivatives across a current carrying surface. The tangential components of the vector potential must be continuous, while the normal components must have a jump due to the presence of a surface current. This can be derived from Ampere's Law and the Coulomb-gauge constraint.
precise
Please refer to the problem towards the end of page in the following link. It's related to discontinuity in normal derivative of magnetic vector potential across a current carrying surface. Prob 5.32 in Griffths.

http://physicspages.com/2013/04/08/magnetostatic-boundary-conditions/

The solution says "Since A is continuous across the surface, the derivatives in directions perpendicular to the surface (that is, {x} and {y}) will be the same on both sides."

I couldn't understand the above argument which says the derivatives along x and y are same across the boundary. What is the calculus I'm missing. Could you explain in detail, or suggest some reading for that math part I'm missing.

I found this thread, with same concern. It's still not clear why though.

Thanks

Last edited:
While your weblink is a bit handwaving to my taste, it's very clearly written in Griffiths's book. The magnetic field has only a finite jump, not a more severe singularity. From
$$\vec{B}=\vec{\nabla} \times \vec{A}$$
you get for any surface $F$ with boundary $\partial F$.
$$\int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}=\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{A}.$$
By the usual arguments with an infinitesimal rectangle penetrating the surface perpendicularly and from the finiteness of $\vec{B}$ along this rectangle you get that in the limit of a vanishing height the surface integral is 0 and so also the line integral over the vector potential. This implies that the tangential components of the vector potential must be continuous.

From the Coulomb-gauge constraint
$$\vec{\nabla} \cdot \vec{A}=0$$
it follows in a similar way by using Gauß's integral theorem that also the normal components of the vector potential are continuous across the surface.

The tangential components of the magnetic field, however must have a jump if a surface current is present as in Griffiths's example. This implies that the derivatives of $\vec{A}$ must have these jumps since $\vec{B}=\vec{\nabla} \times \vec{A}$.

Now I'd go along as follows. Write down Ampere's Law for the above described infinitesimal square, which you orient such that its surface-normal vector is in direction of the surface current $\vec{K}$. Then you get
$$\mu I=\int_F \mathrm{d}^2 \vec{x} (\vec{\nabla} \times \vec{B})=-\int_F \mathrm{d}^2 \vec{x} \cdot \vec{\nabla}^2 \vec{A}.$$
This will immediately give
$$(\vec{n} \cdot \vec{\nabla}) \vec{A}=-\mu_0 \vec{K}.$$
You just have to work out the last integral above in the limit of an infinitesimal square!

1 person

## 1. What is the Magnetic Vector Potential?

The Magnetic Vector Potential is a mathematical quantity used in electrodynamics to describe the magnetic field of a given system. It is denoted by the symbol A and is related to the magnetic field B by the equation B = ∇ × A.

## 2. What is the significance of problem 5.32 in Griffiths' book?

Problem 5.32 in Griffiths' book "Introduction to Electrodynamics" is a challenging problem that involves finding the magnetic vector potential for a given current distribution. It is an important concept in understanding the fundamentals of electrodynamics and is frequently used in many advanced physics and engineering applications.

## 3. How is the magnetic vector potential calculated?

The magnetic vector potential is calculated using the equation A = μ₀/4π ∫(J/|r-r'|)dτ, where μ₀ is the permeability of free space, J is the current density, r is the position where the magnetic field is being calculated, and r' is the position of the current element dτ.

## 4. What is the physical interpretation of the magnetic vector potential?

The physical interpretation of the magnetic vector potential is that it describes the potential energy of a magnetic dipole in a given magnetic field. It also helps in understanding the magnetic forces and interactions between different current-carrying elements in a system.

## 5. How is the magnetic vector potential related to the scalar potential?

The magnetic vector potential and the scalar potential are both components of the electromagnetic potential, which describes the electric and magnetic fields in a given system. The relationship between the two is given by the equation A = -∇φ, where φ is the scalar potential.

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