Magnetic field on rotating electron.

[erhyde]

Homework Statement

an electron is rotating around a proton with a velocity of 2*10^6 m/s on circular path of diameter 5*10^-11 m. what is the magnetic field (B) ?
the centripetal force on the electron and the magnetic force should be equal and mv^2/r = Bqv or
B= mv/qr.

Homework Equations

centripetal force F=mv^2/r
magnetic force = Bqv.

 

[wzy7178]

B=4.55*10^5 t

 

[erhyde]

B=4.55*10^5 t

[STRIKE]can you explain a bit more?[/STRIKE]

nvm found the answer.

 

[erhyde]

new problem

a straight current carrying wire form a loop along its way. what is the magnetic field at the loops centre?
if the loop is rotated 90° about AB axis what will be the new magnetic field?
(sorry for the clumsy pic arrow shows direction of current. )
http://i1367.photobucket.com/albums/r785/erhyde/Untitled_zps91353383.jpg.

 

[paranormal]

I would like to clarify that the above equation for the magnetic field (B) is only valid for a charged particle moving in a straight line. In this scenario, the electron is moving in a circular path, which means the direction of its velocity is constantly changing. Therefore, the magnetic force on the electron is not constant and cannot be equated to the centripetal force.

To accurately calculate the magnetic field in this situation, we would need to use the Lorentz force equation, which takes into account the changing direction of the electron's velocity. This equation is given by F = qvB, where F is the magnetic force, q is the charge of the particle, v is the velocity, and B is the magnetic field.

In order to calculate the magnetic field, we would need to know the charge of the electron and the angle between its velocity and the magnetic field. Without this information, it is not possible to accurately determine the magnetic field in this scenario.