Q.Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at $30^\circ$ to the field direction. Name the trajectories followed by them.

Charged Particle in a Magnetic Field

When a charged particle moves through a magnetic field, the field grabs it sideways. Unlike an electric field, which can speed a charge up or slow it down, a magnetic field only bends the path — it never changes the particle's speed. Understanding why leads directly to circular and helical motion, the basis of cyclotrons, mass spectrometers and the aurora.

The force: always sideways

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the magnetic (Lorentz) force:

Because of the cross product, $\vec{F}$ is perpendicular to both $\vec{v}$ and $\vec{B}$. Its magnitude is

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

Case 1: velocity perpendicular to the field → a circle

If $\vec{v} \perp \vec{B}$ ($\theta = 90^\circ$), the force $F = qvB$ stays constant in size and always points toward one central point. That is exactly the condition for uniform circular motion, with the magnetic force acting as the centripetal force:

Solving for the radius:

The time period of one revolution is

Case 2: velocity at an angle → a helix

If $\vec{v}$ makes an angle $\theta$ with $\vec{B}$, split it into two parts:

• Perpendicular component $v_\perp = v\sin\theta$ — feels the magnetic force and drives circular motion of radius $r = \dfrac{mv_\perp}{qB}$.
• Parallel component $v_\parallel = v\cos\theta$ — feels no force (since $\vec{v_\parallel}\times\vec{B}=0$) and carries the particle steadily along the field line.

Combining a circle with a steady drift gives a helix. The distance advanced along the field in one full turn is the pitch:

A quick example

An electron ($m = 9.1\times10^{-31}\ \text{kg}$, $q = 1.6\times10^{-19}\ \text{C}$) enters a $0.02\ \text{T}$ field at $10^{6}\ \text{m/s}$, perpendicular to $\vec{B}$:

$r = \frac{mv}{qB} = \frac{(9.1\times10^{-31})(10^{6})}{(1.6\times10^{-19})(0.02)} \approx 2.8\times10^{-4}\ \text{m}$ …