- #1

- 57

- 5

- Homework Statement
- Consider the vector potential

A = C ln( (x^2+y^2)/z^2 ) \hat{z}

a) Determine the B-field for this potential. Describe a situation that gives a B-field of that shape.

b) Calculate \nabla \cdot A

c) Give a vector potential in coulomb gauge, A’, which gives the same B as in the a-task.

- Relevant Equations
- B = \nabla x A

My solution for the vector potential ##A=2Cln\frac{x^2+y^2}{z^2} \hat{z}## is:

a) I used the following formula to calculate the magnetic field

$$ \mathbf{B} = \nabla \times \mathbf{A} = \left( \frac{dA_z}{dy} - 0 \right) \hat{x} + \left( 0 - \frac{dA_z}{dx} \right)\hat{y} + 0 \hat{z} = \frac{dA_z}{dy} - \frac{dA_z}{dx} = 2C \frac{x}{x^2+y^2} \hat{x} - 2C \frac{y}{x^2+y^2} \hat{y}. $$

How should I describe a situation with this magnetic field here?

b) I just did the calculation of the formula and got a non-zero value:

$$ \nabla \cdot \mathbf{A} = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right) \cdot \mathbf{A} = \frac{d\mathbf{A}}{dx} + \frac{d\mathbf{A}}{dy} + \frac{d\mathbf{A}}{dz} = 2C \left( \frac{y}{x^2+y¨2} + \frac{x}{x^2+y^2} - \frac{1}{z} \right) $$

c) I was having problem solving this part, I was thinking about using the same formula in question a) ##\mathbf{B} = \nabla \times \mathbf{A'}## and solve for ##\mathbf{A'}## since I have a known ##\mathbf{B}##, but then I realized it is too difficult to solve it. I also thought mathematically that this expression ##\mathbf{A'} = Cln(x^2+y^2)\hat{z}## could give us the same value of B as in the a-task. But this is still not solving the problem correctly. What should I do?

a) I used the following formula to calculate the magnetic field

$$ \mathbf{B} = \nabla \times \mathbf{A} = \left( \frac{dA_z}{dy} - 0 \right) \hat{x} + \left( 0 - \frac{dA_z}{dx} \right)\hat{y} + 0 \hat{z} = \frac{dA_z}{dy} - \frac{dA_z}{dx} = 2C \frac{x}{x^2+y^2} \hat{x} - 2C \frac{y}{x^2+y^2} \hat{y}. $$

How should I describe a situation with this magnetic field here?

b) I just did the calculation of the formula and got a non-zero value:

$$ \nabla \cdot \mathbf{A} = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right) \cdot \mathbf{A} = \frac{d\mathbf{A}}{dx} + \frac{d\mathbf{A}}{dy} + \frac{d\mathbf{A}}{dz} = 2C \left( \frac{y}{x^2+y¨2} + \frac{x}{x^2+y^2} - \frac{1}{z} \right) $$

c) I was having problem solving this part, I was thinking about using the same formula in question a) ##\mathbf{B} = \nabla \times \mathbf{A'}## and solve for ##\mathbf{A'}## since I have a known ##\mathbf{B}##, but then I realized it is too difficult to solve it. I also thought mathematically that this expression ##\mathbf{A'} = Cln(x^2+y^2)\hat{z}## could give us the same value of B as in the a-task. But this is still not solving the problem correctly. What should I do?