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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Motion of a charged particle in uniform magnnetic field(I) is considered one the most difficult concept.

• 49 Questions around this concept.

## Solve by difficulty

If an electron and a proton having the same momenta enter perpendicular to a magnetic field, then

A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is

A charged particle moves through a magnetic field perpendicular to its direction. Then

The period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_{p},\; r_{d},\; and\; r_{\alpha }$ . Which one of the following relations is correct?

## Concepts Covered - 2

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

Motion of a charged particle in uniform magnnetic 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 $F_c=\frac{mv^2}{r}$ . Noting that the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is reduced to $F=qvB$. 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 a 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 traveled (the circumference) divided by the speed. Based on this and 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 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 magnnetic field(II)

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

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

$F_c=\frac{mv^2}{r}$

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

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:

$\begin{array}{l}{v_{p e r p}=v \sin \theta} \\ {v_{p a r a}=v \cos \theta}\end{array}$

where θ 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:

$p=v_{p a r a} T$

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 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 traveling from a very light, thin string to a hard wall and reflecting backward. If the reflection happens at both ends, the particle is trapped in a so-called magnetic bottle.

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

Time period of helical path

$T=\frac{2\pi m}{qB}$

Frequency of helical path

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

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

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

## Study it with Videos

Motion of a charged particle in uniform magnnetic field(I)
Motion of a charged particle in uniform magnnetic field(II)

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