- #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.

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.