Problems on magnetic vector potential

In summary, the magnetic potential, ##\vec A##, must be in the direction of the current, which is in the ##\hat z## direction in cylindrical coordinates. It is only dependent on ##s## and can be represented as $$\vec A = A(s) \hat z$$. The curl of ##\vec A## is equal to the magnetic field, and based on the equation ##\frac{\partial}{\partial s} A(s) = -\frac{\mu_o I s}{2\pi R^2}##, we can see that a negative sign is added due to the use of the curl operator in cylindrical coordinates. This is derived from the formula for the del operator in cylindrical coordinates.
  • #1
Tony Hau
101
30
Homework Statement
Find the magnetic potential inside an infinitely long wire, if it has radius ##R## and the current is uniformly distributed.
Relevant Equations
The magnetic field inside an infinitely long wire is: ##\vec B = \frac{\mu_o I s^2}{R^2} \vec \phi, \text{ }where \text{ }R \gt s##
The magnetic potential outside an infinitely long wire is ##\vec A = -\frac{\mu_o I}{2 \pi}ln(\frac{s}{R})\hat z##, for ##s \geq R##.
The direction of the magnetic potential, ##\vec A##, must be in the direction of the current, which is in ##\hat z## direction in cylindrical coordinates.

It is obvious that the potential only varies with ##s##.

Therefore, $$\vec A = A(s) \hat z$$

Therefore, $$\nabla \times \vec A = \vec B$$ $$\frac{\partial}{\partial s} A(s) = -\frac{\mu_o I s}{2\pi R^2} \text{ } - (Eq \text{ }1)$$ $$A(s)= -\int \frac{\mu_o I s}{2\pi R^2}ds$$ $$= -\frac{\mu_o I}{4\pi R^2}(s^2 -b^2)$$
Here, b is arbitrary, except that since ##\vec A## must be continuous at ##R##.
Therefore, $$-\frac{\mu_o I}{2 \pi}ln(\frac{R}{a}) = -\frac{\mu_o I}{4\pi R^2}(R^2-b^2)$$
It means that we must pick ##a## and ##b## such that ##2ln(\frac{R}{b}) = 1- (\frac{b}{R})^2##
Therefore, $$ a = b = R$$
Therefore, $$ \vec A = -\frac{\mu_o I}{4 \pi R^2}(s^2-R^2) \hat z, \text{ }for \text{ } s \leq R$$.

This is the solution I have copied from the solution manual. One thing I do not understand; why would they add a negative sign at equation 1? Isn't ##\nabla \times \vec A = \vec B##? I know ##\vec E = -\nabla V##, but this applies only to electric potential.
 
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Related to Problems on magnetic vector potential

1. What is magnetic vector potential?

Magnetic vector potential is a mathematical concept used to describe the magnetic field in a given region of space. It is a vector quantity that represents the direction and strength of the magnetic field at a particular point.

2. Why is the magnetic vector potential important?

The magnetic vector potential is important because it helps us understand and analyze the behavior of magnetic fields. It is also a crucial component in the equations that describe the interactions between electric currents and magnetic fields.

3. How is the magnetic vector potential calculated?

The magnetic vector potential is calculated using the Biot-Savart law, which states that the magnetic field at a point is directly proportional to the current flowing through a nearby wire and inversely proportional to the distance from the wire.

4. What are some common problems involving magnetic vector potential?

Some common problems involving magnetic vector potential include calculating the magnetic field at a point due to a current-carrying wire or a solenoid, determining the force on a charged particle in a magnetic field, and analyzing the behavior of magnetic materials.

5. How is the magnetic vector potential related to the electric vector potential?

The magnetic vector potential and the electric vector potential are related through the Maxwell's equations, which describe the fundamental laws of electromagnetism. They are also related through the Lorentz force law, which describes the force on a charged particle in an electromagnetic field.

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