Question

Explain the motion of a charged particle in a uniform magnetic field. Prove that the radius of the path of a charged particle moving perpendicular to a uniform magnetic field is proportional to the momentum of the particle.

Hint:

The magnetic force is perpendicular to velocity, giving circular motion. Set \(qvB = mv^2/r\) and use \(p=mv\).

Answer 1

Step 1: Force on a moving charge.
A charge \(q\) moving with velocity \(v\) in a uniform magnetic field \(B\) experiences the magnetic (Lorentz) force
\[ F = qvB\sin\theta \]
where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\).

Step 2: Particle moving perpendicular to the field.
When the particle moves perpendicular to the field, \(\theta = 90^\circ\) so \(\sin\theta = 1\) and
\[ F = qvB \]
The direction of this force (given by \(\vec{F}=q\,\vec{v}\times\vec{B}\)) is always perpendicular to the velocity. A force that is always perpendicular to the velocity does no work; it only changes the direction of motion, not the speed.

Step 3: Nature of the path.
Since the magnitude of velocity stays constant and the force is constant in magnitude and always perpendicular to \(v\), the particle moves in a circle at constant speed. The magnetic force supplies the necessary centripetal force.

Step 4: Equate magnetic force to centripetal force.
\[ qvB = \frac{mv^2}{r} \]
Cancel one \(v\) from both sides and solve for \(r\):
\[ r = \frac{mv}{qB} \]

Step 5: Relation with momentum.
The linear momentum of the particle is \(p = mv\). Substituting,
\[ r = \frac{p}{qB} \]
For a given particle in a given field, \(q\) and \(B\) are constants, therefore
\[\boxed{r \propto p}\]
Thus the radius of the circular path is directly proportional to the momentum of the charged particle. (The time period \(T = \dfrac{2\pi m}{qB}\) is independent of the speed.)