Force on Moving Charge in a Magnetic Field
Let us consider a positive electric charge 'q' that moves through a magnetic field (magnetic field strength B) with a velocity 'v' and makes an angle '$\theta$' with the magnetic field. Then, it experiences a force due to the field perpendicular to the velocity and the magnetic field.
Experimentally, we can observe that
1) Magnetic force is directly proportional to the magnitude of the charge
$\rm F \propto q$
2) Magnetic force is directly proportional to the velocity of the charge
$\rm F \propto v$
3) Magnetic force is directly proportional to the strength of the magnetic field
$\rm F \propto B$
4) Magnetic force is directly proportional to the sine of the angle between v and B
$\rm F \propto sin \theta$
From the above proportionalities, we get,
$\rm F \propto q v B \sin \theta$
$\rm or, F = k qv B \sin \theta$
where k is a proportionality constant and its value is 1 in SI units. So, we have force,
$\rm F = qvB \sin \theta$
In vector form,
$\rm F = q ( \vec{v} \times \vec{B} )$
Conditions of Force on a Moving Charge in a Magnetic Field
- If the velocity is parallel or antiparallel ($\theta = 0 or \theta = 180$ to the magnetic field, the field exerts a maximum magnetic force on the charge.
- If the velocity is perpendicular to the magnetic field, the field exerts no force on the charge.
- If $\rm v = 0$, then the particle experiences no force due to the magnetic field. This means only charges in motion experience a magnetic force.
- If $\rm q = 0$, then the particle experiences no force due to the magnetic field. This means electrically neutral particles do not observe a magnetic force.
0 Comments
You can let us know your questions in the comments section as well.