 #1
Gustav
 57
 5
 Homework Statement:

Consider the vector potential
A = C ln( (x^2+y^2)/z^2 ) \hat{z}
a) Determine the Bfield for this potential. Describe a situation that gives a Bfield 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 atask.
 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 nonzero 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 atask. 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 nonzero 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 atask. But this is still not solving the problem correctly. What should I do?