 #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, $$\frac{\mu_o I}{2 \pi}ln(\frac{R}{a}) = \frac{\mu_o I}{4\pi R^2}(R^2b^2)$$
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, $$ \vec A = \frac{\mu_o I}{4 \pi R^2}(s^2R^2) \hat z, \text{ }for \text{ } s \leq R$$.
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.
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, $$\frac{\mu_o I}{2 \pi}ln(\frac{R}{a}) = \frac{\mu_o I}{4\pi R^2}(R^2b^2)$$
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, $$ \vec A = \frac{\mu_o I}{4 \pi R^2}(s^2R^2) \hat z, \text{ }for \text{ } s \leq R$$.
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.