- #1
Lambda96
- 206
- 72
- Homework Statement
- Calculate the electric and magnetic field
- Relevant Equations
- none
Hi
I have problems with the following task
I now wanted to try to calculate the vector potential, which according to my professor's script is defined as follows:
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$
I have now visualized the problem as follows in 2D with regard to the x and z axes
I'm not so sure, but does the electric as well as the magnetic field consist of a superposition when the particle is at the position ##z_1=a## and ##z_2=-a## and would look like ##E=E_1+E_2## and ##B=B_1+B_2##?
Then the following ##\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)## and ##\mathbf{x'}_2=\left(\begin{array}{c} 0 \\\ 0 \\ -a \end{array}\right)## would hold and thus ##| \mathbf{x} -\mathbf{x'}_1 |=\sqrt{x^2+y^2+(z-a)^2}##
The electrical current density is defined as follows ##\mathbf{j}=\rho \mathbf{v}## and for this task:
$$\mathbf{j} = Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y) \cdot v \mathbf{e}_z$$
The vector potential can then be calculated as follows:
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d^3\mathbf{x}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$
Would now simplify the potential because ##\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)## as follows
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$
I have problems with the following task
I now wanted to try to calculate the vector potential, which according to my professor's script is defined as follows:
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$
I have now visualized the problem as follows in 2D with regard to the x and z axes
I'm not so sure, but does the electric as well as the magnetic field consist of a superposition when the particle is at the position ##z_1=a## and ##z_2=-a## and would look like ##E=E_1+E_2## and ##B=B_1+B_2##?
Then the following ##\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)## and ##\mathbf{x'}_2=\left(\begin{array}{c} 0 \\\ 0 \\ -a \end{array}\right)## would hold and thus ##| \mathbf{x} -\mathbf{x'}_1 |=\sqrt{x^2+y^2+(z-a)^2}##
The electrical current density is defined as follows ##\mathbf{j}=\rho \mathbf{v}## and for this task:
$$\mathbf{j} = Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y) \cdot v \mathbf{e}_z$$
The vector potential can then be calculated as follows:
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d^3\mathbf{x}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$
Would now simplify the potential because ##\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)## as follows
$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$