How Does the Magnetic Vector Potential Change with Infinite Wire Length?

[lovinuz]

Homework Statement

A finite wire lies on the z axis and extends from the point z=-V to z=+V. the equation from griffiths introduction to electrodynamics (equation 5.62) can be used to determine the vector potential at a point in the xy plane a distance s from the wire to be A=mu(naught)I/4pi [integral of dx'/sqrt(s square + x' square) from -V to +V] k, where k is the direction vector. after calculation, it gives: A=mu(naught)I/2pi ln(V/s + sqrt[1+(V/s)square] k. for a finite wire, we should take the limit as V goes to infinity, but the result blows up in this case. the first part of my question is to show that in the limit V goes to infinity, the above result becomes A=mu(naught)I/2pi ln(2V/s) k this is simple as we can take V/s common in the logarithm part, and then take the limit of V to infinity to get the required equation. the second part of the question asks to take the curl of the achieved equation using cylindrical coordinates, and thereby determining the magnetic field B at points in the xy plane. this is where i get stuck!
the third part is even harder and gives us that an equally good vector potential is given by A'=A+(delta)(lambda), where lambda is any scalar field we might wish to choose. what i need is a suitable choice for lambda so that A' remains finite in the limite V goes to infinity.

these two parts are giving me a hard time, i am sorry if i was very untidy in my representation, but if anyone could help me it wud be really nice, thanks!

 

[jbsteele]

Homework EquationsGriffiths equation 5.62A=mu(naught)I/4pi [integral dx'/sqrt(s square + x' square) from -V to +V] kThe Attempt at a SolutionThe attempt at a solution for the second part is as follows: We can use the cylindrical coordinate system to take the curl of the vector potential, A. First, we expand A in terms of its components in the r and z directions: A = (mu(naught)I/2pi ln(2V/s))r/s + (mu(naught)I/2pi ln(2V/s))z/s Then, we take the curl of A using the following identities: curl(A) = (1/r)(∂Az/∂θ) - (1/r)(∂Ar/∂z)Substituting the components of A into the above equation, we get: curl(A) = (mu(naught)I/2π)(1/rs)(1/s)(1/2V)(2V/s) - (mu(naught)I/2π)(1/rs)(-1/s)(1/2V)(2V/s)Which simplifies to: curl(A) = 0 Therefore, the magnetic field B is zero at points in the xy plane. For the third part, we start by expanding A' in terms of its components in the r and z directions: A' = (mu(naught)I/2π ln(2V/s))r/s + (mu(naught)I/2π ln(2V/s))z/s + (delta)λ Now, we can choose λ such that A' remains finite in the limit as V goes to infinity. For this, we choose λ = -mu(naught)I/2π ln(2V/s). Substituting this into the above equation, we get: A' = (delta)(-mu(naught)I/2π ln(2

 

[thomate1]

I would like to provide a response to the content regarding the magnetic vector potential. The equation provided from Griffiths' Introduction to Electrodynamics (equation 5.62) is a well-known equation used to determine the vector potential at a point in the xy plane a distance s from a finite wire lying on the z axis. The equation is derived from the Biot-Savart law and is a crucial tool in understanding the behavior of magnetic fields.

The first part of the content asks to show that in the limit V goes to infinity, the result of the equation becomes A=mu(naught)I/2pi ln(2V/s) k. This is a straightforward calculation, as taking the limit of V to infinity will result in the term [sqrt(1+(V/s)^2)] becoming simply 1, and therefore the equation simplifies to A=mu(naught)I/2pi ln(2V/s) k.

The second part of the content asks to take the curl of the achieved equation using cylindrical coordinates, and determine the magnetic field B at points in the xy plane. This is where I believe the confusion lies. The equation given is for the vector potential, not the magnetic field. To obtain the magnetic field B, we need to use the relation B= curl(A). In this case, the curl of the equation A=mu(naught)I/2pi ln(2V/s) k would be zero, as the vector potential is constant in the z-direction and has no dependence on the cylindrical coordinates. Therefore, the magnetic field B would be zero at points in the xy plane.

The third part of the content mentions that an equally good vector potential can be given by A'=A+(delta)(lambda), where lambda is any scalar field. The choice of lambda is crucial in ensuring that A' remains finite in the limit V goes to infinity. One possible choice for lambda could be a function that decreases to zero as V goes to infinity, such as lambda=1/V. This would result in A' approaching the same value as A in the limit V goes to infinity, ensuring that A' remains finite.

In summary, the concept of magnetic vector potential is a powerful tool in understanding the behavior of magnetic fields. However, it is important to note that the equations for the vector potential and the magnetic field are not interchangeable, and care must be taken when using them. Additionally, the