Let us start with a new problem that involves an imaginary magnetic monopole, producing magnetic field that follows inverse square law similar to the electrostatic case.
As a consequence, we can exploit some more information about this system.
A combination of electric monopole and a magnetic monpole is known to be Thomson dipole.
Here, we assume the magnetic field of the magnetic charge is equal to , $$ \vec{B} = \frac{q_m}{r^2}\hat{r}$$ Just from the Lorentz force, it is known that Magnetic force can never do work on a electric charge since it is always perpendicular to the velocity. Similarly, the electric force can never do work on magnetic charge.
Mathematically, work done by the Magnetic field (of the magnetic charge) on electric charge is, $$ dW = \int \vec{F_{em}}\cdot\vec{dl} = q_e \int(\frac{\vec{v}}{c}\times\vec{B})\cdot\vec{dl} \\~\\ = q_e \int (\vec{v}\times\vec{B})\cdot \vec{v}\, dt = 0 $$
Similarly, work done by Electric field on Magnetic charge is always zero, $$dW = \int\vec{F_{me}} \cdot\vec{dl} = \frac{-\vec{v}}{c}\times\vec{E}\cdot\vec{v}\, dt = 0 $$
From this fact, we can be sure that the particles will not gain any kinetic energy from Work-Energy theorem. So, the velocity of the particles should be a constant.
Other than this, you can also determine the basic constants of motion you can find in any mechanics problem i.e.Total Energy, total linear and angular momentum.
Except now, the definition of momentum is upgraded to Electromagnetic momentum which has wider applicability than the former definition.
There are some astonishing results about this system because of the above fact.
First, if you solve the system completely you will find out that the motion of the particle will lie only along the surface of cone centered about an axis passing through the direction of a constant vector quantity i.e. $$\vec{L'} = (\vec{r}\times{m\vec{v}}) - q_eq_m\hat{r} $$.
where the spherical coordinates are measured from where the z-axis lie along the direction of vector Q.
And the Second which is very peculiar that,
this system has some intrinsic angular momentum stored in its fields which is independent of the distance between the charges. This in fact happens to be the basis for the idea of quantization of the Electric and Magnetic charge in Quantum mechanics - one of the famous ideas of Dirac.
It is derived as follow,
The magnetic field, $$ \vec{B} = \frac{q_m}{r^3}\vec{r}$$ and the electric field is placed at a distance "d" from this origin. So,
the Electric field, $$ \vec{E} = \frac{q_e}{r'^3} \vec{r'} $$
Using vector addition rule, $$ \vec{r'} = \vec{r} + \vec{d} $$ where $\vec{d}$ is directed from electric charge to magnetic charge.
Henceforth, $$ \vec{E} = \frac{q_e(\vec{r}+\vec{d})}{(r^2 + d^2+ 2\,r\,d\, cos\theta)^{3/2}} $$ Then linear momentum density stored in the fields is calculated as, $$ \vec{P} = \frac{1}{4\pi{c}}\vec{E}\times\vec{B}= \frac{q_eq_m}{4\pi{c}}\frac{((\vec{r}+\vec{d})\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}\\~\\ = \frac{q_eq_m}{4\pi{c}}\frac{(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
Now, finding the angular momentum density stored in the fields,
$$ \vec{l} = \vec{r}\times\vec{P} = \frac{q_eq_m}{4\pi{c}}\frac{\vec{r}\times(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
Using the vector relation, $$ \vec{r}\times(\vec{d}\times\vec{r}) = \vec{d} (\vec{r}\cdot\vec{r}) - \vec{r} (\vec{r}\cdot\vec{d}) $$
$$ \vec{l} = \frac{q_eq_m}{4\pi{c}}\frac{(r^2 \vec{d} - r\,d\,cos\theta\, \vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
To get the total angular momentum, we integrate this all over the space using spherical coordinate system where we assume $r\,cos\theta$ lie along the direction $\vec{d}$ and we can split the vector into its components as $\vec{r} = r cos\theta \hat{d} + \,\,components\,\, perpendicular\,\, to\,\, the \,\,direction\,\,\hat{d}$$
so that $$ \vec{L} = \frac{q_eq_m}{4\pi{c}}\int_{space}\frac{(r^2 \,d - r^2\,d\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} d\tau $$
The other perpendicular components will integrate to the value zero. $$ \vec{L} = \frac{q_eq_md}{4\pi{c}}\int_{space}\frac{(r^2 - r^2\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} r^2 sin\theta \,d\theta \,d\phi\, dr $$
On integration we will get, $$ \vec{L} = \frac{q_eq_m}{c} $$
When we go to quantum mechanics, we have studied the angular momentum is quantized in terms $$ L = n\frac{\hbar}{2} $$
Comparing the results we get, $$ L = \frac{q_eq_m}{c} = n\frac{\hbar}{2} $$
Therefore, $$ \frac{2q_eq_m}{\hbar{c}} = Integer $$
which is the exact result obtained by Dirac. From this condition, even if one magnetic charge exists in nature, it would imply the quantization of all the electric charges in the Universe.
As a consequence, we can exploit some more information about this system.
A combination of electric monopole and a magnetic monpole is known to be Thomson dipole.
Here, we assume the magnetic field of the magnetic charge is equal to , $$ \vec{B} = \frac{q_m}{r^2}\hat{r}$$ Just from the Lorentz force, it is known that Magnetic force can never do work on a electric charge since it is always perpendicular to the velocity. Similarly, the electric force can never do work on magnetic charge.
Mathematically, work done by the Magnetic field (of the magnetic charge) on electric charge is, $$ dW = \int \vec{F_{em}}\cdot\vec{dl} = q_e \int(\frac{\vec{v}}{c}\times\vec{B})\cdot\vec{dl} \\~\\ = q_e \int (\vec{v}\times\vec{B})\cdot \vec{v}\, dt = 0 $$
Similarly, work done by Electric field on Magnetic charge is always zero, $$dW = \int\vec{F_{me}} \cdot\vec{dl} = \frac{-\vec{v}}{c}\times\vec{E}\cdot\vec{v}\, dt = 0 $$
From this fact, we can be sure that the particles will not gain any kinetic energy from Work-Energy theorem. So, the velocity of the particles should be a constant.
Other than this, you can also determine the basic constants of motion you can find in any mechanics problem i.e.Total Energy, total linear and angular momentum.
Except now, the definition of momentum is upgraded to Electromagnetic momentum which has wider applicability than the former definition.
There are some astonishing results about this system because of the above fact.
First, if you solve the system completely you will find out that the motion of the particle will lie only along the surface of cone centered about an axis passing through the direction of a constant vector quantity i.e. $$\vec{L'} = (\vec{r}\times{m\vec{v}}) - q_eq_m\hat{r} $$.
where the spherical coordinates are measured from where the z-axis lie along the direction of vector Q.
And the Second which is very peculiar that,
this system has some intrinsic angular momentum stored in its fields which is independent of the distance between the charges. This in fact happens to be the basis for the idea of quantization of the Electric and Magnetic charge in Quantum mechanics - one of the famous ideas of Dirac.
It is derived as follow,
The magnetic field, $$ \vec{B} = \frac{q_m}{r^3}\vec{r}$$ and the electric field is placed at a distance "d" from this origin. So,
the Electric field, $$ \vec{E} = \frac{q_e}{r'^3} \vec{r'} $$
Using vector addition rule, $$ \vec{r'} = \vec{r} + \vec{d} $$ where $\vec{d}$ is directed from electric charge to magnetic charge.
Henceforth, $$ \vec{E} = \frac{q_e(\vec{r}+\vec{d})}{(r^2 + d^2+ 2\,r\,d\, cos\theta)^{3/2}} $$ Then linear momentum density stored in the fields is calculated as, $$ \vec{P} = \frac{1}{4\pi{c}}\vec{E}\times\vec{B}= \frac{q_eq_m}{4\pi{c}}\frac{((\vec{r}+\vec{d})\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}\\~\\ = \frac{q_eq_m}{4\pi{c}}\frac{(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
Now, finding the angular momentum density stored in the fields,
$$ \vec{l} = \vec{r}\times\vec{P} = \frac{q_eq_m}{4\pi{c}}\frac{\vec{r}\times(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
Using the vector relation, $$ \vec{r}\times(\vec{d}\times\vec{r}) = \vec{d} (\vec{r}\cdot\vec{r}) - \vec{r} (\vec{r}\cdot\vec{d}) $$
$$ \vec{l} = \frac{q_eq_m}{4\pi{c}}\frac{(r^2 \vec{d} - r\,d\,cos\theta\, \vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$
To get the total angular momentum, we integrate this all over the space using spherical coordinate system where we assume $r\,cos\theta$ lie along the direction $\vec{d}$ and we can split the vector into its components as $\vec{r} = r cos\theta \hat{d} + \,\,components\,\, perpendicular\,\, to\,\, the \,\,direction\,\,\hat{d}$$
so that $$ \vec{L} = \frac{q_eq_m}{4\pi{c}}\int_{space}\frac{(r^2 \,d - r^2\,d\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} d\tau $$
The other perpendicular components will integrate to the value zero. $$ \vec{L} = \frac{q_eq_md}{4\pi{c}}\int_{space}\frac{(r^2 - r^2\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} r^2 sin\theta \,d\theta \,d\phi\, dr $$
On integration we will get, $$ \vec{L} = \frac{q_eq_m}{c} $$
When we go to quantum mechanics, we have studied the angular momentum is quantized in terms $$ L = n\frac{\hbar}{2} $$
Comparing the results we get, $$ L = \frac{q_eq_m}{c} = n\frac{\hbar}{2} $$
Therefore, $$ \frac{2q_eq_m}{\hbar{c}} = Integer $$
which is the exact result obtained by Dirac. From this condition, even if one magnetic charge exists in nature, it would imply the quantization of all the electric charges in the Universe.
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