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: Set up the geometry.
Let a particle of charge \(q\), mass \(m\) and speed \(v\) enter a uniform field \(B\) at right angles to it. The field does not act along the motion, so the particle is deflected sideways continuously.

Step 2: Use the work-energy idea.
The Lorentz force \(\vec{F}=q\,\vec{v}\times\vec{B}\) is perpendicular to \(\vec{v}\), so \(\vec{F}\cdot\vec{v}=0\). Hence power delivered is zero and the kinetic energy (and speed \(v\)) never changes. Only the direction turns, which means uniform circular motion.

Step 3: Angular form of the balance.
For circular motion of radius \(r\), centripetal force \(= m\omega^2 r\) where \(\omega = v/r\). The magnetic force \(qvB\) provides it:
\(qvB = m\dfrac{v^2}{r}\).

Step 4: Solve for the radius.
\(r = \dfrac{mv}{qB}\). Writing momentum \(p = mv\) gives \(r = \dfrac{p}{qB}\).

Step 5: Read off the proportionality.
With \(q\) and \(B\) fixed, \(r\) changes only through \(p\), so doubling the momentum doubles the radius.
\[\boxed{r = \frac{p}{qB} \ \Rightarrow\ r \propto p}\]