Motion Of A Charged Particle In Uniform Magnnetic Field - Practice Questions & MCQ

Motion of a charged particle in uniform magnetic field(I)

In the figure a negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper. The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion. Note that because the charge is negative, the force is opposite in direction to the prediction of the right-hand rule.

In this situation, the magnetic force supplies the centripetal force $F_c=\frac{m v^2}{r}$. Noting that the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is reduced to $F=q v B$. Because the magnetic force $F$ supplies the centripetal force $F_c$ we have,

$
q v B=\frac{m v^2}{r}
$

Solving for $r$ gives

$
r=\frac{m v}{q B}
$

Here, $r$ is the radius of curvature of the path of a charged particle with mass $m$ and charge $q$, moving at speed $v$ that is perpendicular to a magnetic field of strength $B$. The time for the charged particle to go around the circular path is defined as the period, which is the same as the distance travelled (the circumference) divided by the speed. Based on this and the Equation, we can derive the period of motion as:

$
T=\frac{2 \pi r}{v}=\frac{2 \pi}{v} \frac{m v}{q B}=\frac{2 \pi m}{q B}
$

Therefore, the frequency of revolution is

$
\frac{1}{T}=\frac{q B}{2 \pi m}
$

This frequency is called the cyclotron frequency.

Motion of a charged particle in uniform magnetic field(II)

In the previous topic, we have derived the expression for centripetal force on a charged particle moving in a uniform magnetic field.

$
F_c=\frac{m v^2}{r}
$

The radius of curvature as

$
r=\frac{m v}{q B}
$

If the velocity is not perpendicular to the magnetic field, then we can compare each component of the velocity separately with the magnetic field. The component of the velocity perpendicular to the magnetic field produces a magnetic force perpendicular to both this velocity and the field:

$
\begin{aligned}
& v_{\text {perp }}=v \sin \theta \\
& v_{\text {para }}=v \cos \theta
\end{aligned}
$

where $\theta$ is the angle between $v$ and $B$. The component parallel to the magnetic field creates constant motion along the same direction as the magnetic field, also shown in Equation. The parallel motion determines the pitch $p$ of the helix, which is the distance between adjacent turns. This distance equals the parallel component of the velocity times the period:

$
p=v_{\text {para }} T
$
 

This results in  a helical motion, as shown in the following figure: 

While the charged particle travels in a helical path, it may enter a region where the magnetic field is not uniform. In particular, suppose a particle travels from a region of the strong magnetic field to a region of weaker field, then back to a region of stronger field. The particle may reflect back before entering the stronger magnetic field region. This is similar to a wave on a string travelling from a very light, thin string to a hard wall and reflecting backwards. If the reflection happens at both ends, the particle is trapped in a so-called magnetic bottle.

The radius of helical path

$
r=\frac{m(v \sin \theta)}{q B}
$

Time period of the helical path

$
T=\frac{2 \pi m}{q B}
$

Frequency of helical path

$
F=\frac{1}{T}=\frac{q B}{2 \pi m}
$

Pitch: The pitch is the horizontal distance between two consecutive circles.

$
P=(V \cos \theta) T=\frac{2 \pi m}{q B}(V \cos \theta)
$