Question

19. Consider a charged particle (charge q) at the origin surrounded by an infinitely long cylindrical shell (radius a) of a magnetic field in the φ direction, i.e. B = Ao δ[s- (a) Determine the electromagnetic momentum density g-60 EB. (b) Determine the total momentum in the field by integrating over all space. (c) Determine the vector potential corresponding to the magnetic field and show that the electromagnetic momentum is just the vector potential at the position of the charge time the charge. This result is connected with the fact that mv+q A is the canonical momentum of a particle with mass m and charge q. (d) Show that if the field is suddenly turned off, the charge gains a momentum equal to the momentum in the field.

Answer 1

The electric field of the charge is
  $\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{\vec{r}}{r^3}$
Now, in the cylindrical coordinates, we get
  $\vec{r}=s\hat{s}+z\hat{z}$
And
$r=|\vec{r}|=\sqrt{s^2+z^2}$
So, we get
   $\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{s\hat{s}+z\hat{z}}{(s^2+z^2)^{3/2}}$
And the given magnetic vector field is
   $\vec{B}=A_0\delta (s-a)\hat{\phi}$
So,
  $\vec{g}=\epsilon_0(\vec{E}\times \vec{B})=\frac{q}{4\pi}A_0\delta (s-a)\frac{1}{(s^2+z^2)^{3/2}}(s\hat{s}+z\hat{z})\times \hat{\phi}$
$\Rightarrow \vec{g}=\epsilon_0(\vec{E}\times \vec{B})=\frac{q}{4\pi}A_0\frac{\delta (s-a)}{(s^2+z^2)^{3/2}}(s\hat{z}-z\hat{s})$
$\Rightarrow \vec{g}=\epsilon_0(\vec{E}\times \vec{B})=\frac{q}{4\pi}A_0\frac{\delta (s-a)}{(s^2+z^2)^{3/2}}(s\hat{z}-z(\cos\phi \hat{x}+\sin\phi \hat{y}))$
b)
The total momentum in the field is given by
  $\vec{P}_{EM}=\int \vec{g}d^3 r=\int\epsilon_0(\vec{E}\times \vec{B})d^3 r=\frac{qA_0}{4\pi}\int s ds d\phi dz \frac{\delta (s-a)}{(s^2+z^2)^{3/2}}(s\hat{z}-z(\cos\phi \hat{x}+\sin\phi \hat{y}))$
$\Rightarrow \vec{P}_{EM}=\frac{qA_0}{4\pi}\int s ds ~ dz \frac{\delta (s-a)}{(s^2+z^2)^{3/2}}(2\pi s\hat{z})$
$\Rightarrow \vec{P}_{EM}=\hat{z}\frac{qA_0}{2}\int s^2 ds ~ dz \frac{\delta (s-a)}{(s^2+z^2)^{3/2}}$
$\Rightarrow \vec{P}_{EM}=\hat{z}\frac{qA_0a^2}{2}\int_{-\infty}^{\infty} dz \frac{1}{(a^2+z^2)^{3/2}}$
$\Rightarrow \vec{P}_{EM}=\hat{z}{qA_0a^2}\int_{0}^{\infty} \frac{dz}{(a^2+z^2)^{3/2}}$
$\Rightarrow \vec{P}_{EM}=\hat{z}{qA_0a^2}\frac{1}{a^2}$
$\Rightarrow \vec{P}_{EM}=\hat{z}qA_0$
c)

  The vector potential corresponding to the given magnetic field is calculated as follows:
  As we have
  $\vec{B}=\nabla\times \vec{A}$
So, we get
  $\int_S \vec{B}.\vec{dS}=\int_S (\nabla\times \vec{A}).\vec{dS}=\oint_C \vec{A}.\vec{dl}$
$\Rightarrow \int_S \vec{B}.\vec{dS}=\oint_C \vec{A}.\vec{dl}$
Now we consider a surface area which is perpendicular to the magnetic field B and for simplicity we take it as a rectangle of length along z axis as L and length along cylindrical radial direction by b and this surface is cutting the shell of radius a. So, we get
   $\vec{dS}=\hat{\phi}~ds~dz$
And as we have
  $\vec{B}=A_0\delta (s-a)\hat{\phi}$
from the symmetry of the problem, we have the vector potential along z direction. i.e.,
  $\vec{A}=A\hat{z}$ (inside, s<a ) and A = 0 (for s > a). This is because this theta function will give the delta function when we take the derivative to calculate the B from A.
So, from the flux condition, now, we get
$\Rightarrow \int_S \vec{B}.\vec{dS}=\oint_C \vec{A}.\vec{dl}$
$\Rightarrow A_0\int \delta(s-a)ds\int dz=A\int {dz}$
$\Rightarrow A=A_0$
So, we get
  $\Rightarrow \vec{A} =\hat{z}A=\hat{z}A_0$
So, the electromagnetic momentum is just the vector potential at the position of the charge times the charge of the particle. i.e.,
  $\Rightarrow \vec{P}_{EM}=\hat{z}qA_0=q\vec{A}$
d)

  Initially there was no force on the charge. So, if we assume that the charge is at rest initially (v =0), then, we get
  the canonical momentum of the charge to be
  $\vec{p}_{charge, initial}=m\vec{v}&plus;q\vec{A}=q\vec{A}=qA_0\hat{z}$
Now, as the field is suddenly switched off (A =0, now), so, the final canonical momentum is
  $\vec{p}_{charge, final}=m\vec{v}$
As there is no force on the charge, so, the canonical momentum should be conserved. i.e.,
  $\vec{p}_{charge, final}=\vec{p}_{charge, initial}\Rightarrow m\vec{v}=q\vec{A}=qA_0\hat{z}$
Thus, the charge gains a mechanical momentum equal to the momentum in the field.