How Does the Biot-Savart Law Apply to an Electron Orbiting a Proton?

[dekoi]

Suppose an electron orbits a proton at a distance of 'r' with speed 'v'.
How could i determine the magnitude of the magnetic field at the location of the proton?

I thought it would make sense to use The Biot-Savart Law, but i don't know where to begin:
$$dB = \frac{\mu_o}{4\pi} \frac{I ds x r}{r^2}$$
(where $$ds x r$$ is a cross product of ds and r (unit vector) )

 

[mezarashi]

You'll be doing an integration around the circumference of the orbit. The vectors ds and r will always be perpendicular so you can ignore the vector operation and focus on magnitude.

A helping relationship would be that of current and electron flow. By definition:

J = nqv

where J is the current density, n is the electron density, q is the charge, and v is the average drift velocity. Since you have only one electron, you could probably assume I = qv/C, where C is the orbit's circumference.

 

[quicknote]

The Biot-Savart Law is a fundamental equation in electromagnetism that allows us to calculate the magnetic field at a specific point due to a current-carrying wire or a moving charge. In this case, we have a moving electron orbiting a proton, which can be treated as a current-carrying particle.

To determine the magnitude of the magnetic field at the location of the proton, we can use the Biot-Savart Law in its differential form:

dB = \frac{\mu_o}{4\pi} \frac{I ds x r}{r^2}

where dB is the magnitude of the magnetic field at a point, \mu_o is the permeability of free space, I is the current carried by the electron, ds is a small element of the electron's path, and r is the distance from the element ds to the point where we want to calculate the field.

In this case, we can consider the electron's path as a circular orbit around the proton, with a radius of r. The current carried by the electron can be calculated as I = qv, where q is the charge of the electron and v is its speed. The direction of the current is given by the direction of the electron's motion, which is perpendicular to the radius vector r.

Substituting these values into the Biot-Savart Law, we get:

dB = \frac{\mu_o}{4\pi} \frac{q v ds x r}{r^2}

Now, we can use the right-hand rule to determine the direction of the cross product ds x r. Since the electron is moving in a circular orbit, the direction of ds is tangential to the orbit and the direction of r is radial, pointing towards the proton. Therefore, the cross product will be in the direction perpendicular to both ds and r, which is in the same direction as the magnetic field.

Integrating over the entire orbit, we get:

B = \frac{\mu_o}{4\pi} \frac{q v}{r^2} \int ds

The integral of ds over the entire orbit is equal to the circumference of the circle, which is 2\pi r. Therefore, the final expression for the magnitude of the magnetic field at the location of the proton is:

B = \frac{\mu_o}{2} \frac{q v}{r}

This shows that the magnitude of the magnetic field is directly proportional