find the magnetic field at a distance r from an infinite wire

Question

find the magnetic field at a distance r from an infinite wire through which a stationary current passes I. Then find the magnetic vector potential

Answer 1

Hi Wyzantstudent V,

 

First of all, let's look at the symmetry. Let's use the infinite wire as our z-axis and a plane perpendicular to it - as xy-plane with z=0

Mark that plane as P.

Because the wire extends to infinity both above the P and below the P, there must be no preference between the z-direction above the P and that below the P. Therefore,

1. the vector of magnetic field created by the wire can have only zero z-component. Another words,

1. only components of the magnetic field in a planes perpendicular to the wire are possible

2. In a similar way, b/c the wire is infinite, the field in a plane P must be the same as in any other plane Q parallel shifted above or below P at any z-distance from it.

Another words,

1. components of the field should not depend on z

3. All points in a plane P that are equally distant from the wire (z axis), should be physically equivalent. Therefore,

1. the field at a point (x,y) in a plane P must depend only on the distance ρ from the wire (along the plane)

4. Finally, the magnetic field vectors in a plane P cannot have radial components.

Indeed, if we assume a non-zero radial component of a magnetic field (which according to the symmetry property 3.above must be directing toward the wire (at all points on a circle) or directing at all points away from the wire), such magnetic field would have nonzero flux. But one of the fundamental properties of (static, our current doesn't change in time) magnetic fields is that they have zero flux.

Summarizing, what we have said above:

Magnetic field vectors of an infinite wire must a) lye in planes perpendicular to the wire b) depend only on the distance from the wire c) have field lines in the shape of concentric circles around the point of intersection of the wire with the plane.

All the above is the consequence of the symmetry of the problem's setup, in our case the axial (or cylindrical) symmetry.

I'm solving electro-or-magnetostatics problems, choosing the fitting coordinate system would greatly simplify the solution

In our case, instead of (x, y, z) Cartesian coordinates

it is more appropriate to use the cylindrical (ρ,φ,z) where ρ ,- distance from the z-axis (wire) in a plane Q; φ- angle counted in counter-clock direction from an (once chosen) arbitrary direction from the intersection O of z-axis with Q, z- distance of Q from P along z

After that preparation we could proceed to the calculations. Using the the fact that the magnetic field would have only Bφ, component the integral along a circular field line of radius ρ would become rather easy to take.

After that using ∇xB in cylindrical coordinates, the vector potential would also become fairly simple

A simple way to get the magnetic induction B in 

our symmetric case would be to apply the Stokes theorem to the Maxwell Equation  

  curl B=μ₀j (1)

If we take a surface integral of both sides of (1) over an area of a circle of radius ρ in the plane P

with the circle's center at the point O of intersection of the wire with the plane P, The Stokes theorem would let us replace the surface integral of the l.h.s of (1) with a line integral over the circumference of the circle.

1. Because (as we saw above), the vector B must have non-zero only the component tangent to the circle, Bφ which is independent on the angle φ, the line integral over the circumference would become equal to Bφ x 2πρ . 
2. As to the surface integral of the surface current density in the r.h.s. of (1), it should be equal to μ₀x I
3. Therefore Bφ (2πρ)=μ₀I (2)

or 

Bφ = μ₀I/2πρ (3)

To find the vector potential A - such that 

curl A=B (4)

it would be most convenient (again, only b/c of axial symmetry of the setup) to look at (4) in cylindrical coordinates (ρ,φ,z). Because B has only Bφ non-zero component, we need only φ - component of the l.h.s. of (4). (below the symbol δ is used as a symbol of partial derivative)

(curl A)φ=(δAρ/δz)-(δΑz/δρ) (5)

 Assuming (pls. see above) that no properties should depend on z, we could combine (5) and (3)

 -(δΑz/δρ)=μ₀I/2πρ (6)

or 

δΑz=-(μ₀I/2π)δρ/ρ (7)

which has a solution 

Αz(ρ)=-(μ₀I/2π)lnρ+Const =

(μ₀I/2π)ln(1/ρ)+Const (8)

This is the end of the solution

Hope it was helpful

Dr.Ariel B.