Find a vector potential A^ from a long straight wire carrying current ...
Vector Potential A^ from a Long Straight Wire Carrying Current I
To find the vector potential A^ from a long straight wire carrying current I, we can use the Biot-Savart law:
A^ = (μ/4π) ∫(J(r')/|r-r'|)dτ'
Where μ is the permeability of free space, J(r') is the current density at position r', |r-r'| is the distance between r and r', and dτ' is the volume element at position r'.
For a long straight wire carrying current I, we can assume that the current density is uniform and directed along the wire, so J(r') = Iδ(r'-r), where δ is the Dirac delta function. This simplifies the integral to:
A^ = (μ/4π) I ∫(δ(r'-r)/|r-r'|)dτ'
Since the wire is infinitely long, we can assume that the wire extends to infinity in both directions. This means that we can choose our integration path to be a circle with radius r centered on the wire. Using cylindrical coordinates, we can write:
A^ = (μ/4π) I ∫(δ(ρ-r)/ρ)dρdφdz
The integral over φ and z can be evaluated trivially, leaving us with:
A^ = (μ/4π) I ∫(δ(ρ-r)/ρ)ρdρ
Which simplifies to:
A^ = (μ/4π) I ln(ρ/r)
Magnetic Field from the Vector Potential
The magnetic field can be found from the vector potential using the equation:
B^ = ∇ x A^
Where ∇ is the gradient operator. In cylindrical coordinates, this becomes:
Bρ = (1/ρ)(∂Az/∂φ) - (∂Ay/∂z)
Bφ = (∂Ax/∂z) - (∂Az/∂ρ)
Bz = (1/ρ)(∂Ar/∂φ) - (∂Aφ/∂ρ)
Using the vector potential we found earlier:
A^ = (μ/4π) I ln(ρ/r)
We can find the components of the magnetic field:
Br = 0
Bφ = (μ/4π) I (1/ρ)
Bz = 0
Since the magnetic field only has a φ component, it is directed tangentially to the wire and has a magnitude given by:
|B| = (μ/4π) I (1/ρ)
Conclusion
In summary, we can find the vector potential A^ from a long straight wire carrying current I using the Biot-Savart law. From the vector potential, we can find the magnetic field using the equation B^ = ∇ x A^. For a long straight wire, the magnetic