Motion of a Charged Particle in a Magnetic Field

Motion of a Charged Particle in a Magnetic Field

The magnetic force ${\overrightarrow{\text{F}}}_{\text{m}}$on a charge q moving with velocity $\overrightarrow{\text{v}}$ in a magnetic field $\overrightarrow{\text{B}}$ is given by the expression.

${\vec{F}}_m=q(\vec{v}\times \vec{B})$

Here $\overrightarrow{V}$ = velocity of the particle and

$\overrightarrow{B}$ = magnetic field

On the basis of above expression, we can draw the following conclusion.

(a) Stationary charge (i.e. $\overrightarrow{V}$ = 0) experiences no magnetic force.

(b) If $\overrightarrow{V}$ is parallel or anti parallel to $\overrightarrow{B}$ then the charged particle experiences no magnetic force.

(c) Magnetic force is always perpendicular to both the $\overrightarrow{V}$ and $\overrightarrow{B}$.

(d) As magnetic force is always perpendicular to the $\overrightarrow{V}$, so magnetic force does not deliver power to the charged particle.

(e) As magnetic force is always perpendicular to the $\overrightarrow{V}$, hence this force provides the necessary centripetal force to the charged particle to move on circular path.

On the basis of expression $\vec{F}=q(\vec{v}\times \vec{B})$ the maximum value of magnetic force is equal to

F=qvB, which occurs when the charge is projected perpendicular to the magnetic field. In this case path of charged particle is purely circular (in uniform _{$\overrightarrow{B}$}) and magnetic force provides necessary centripetal force.

(f) If radius of the circular path is R then,

$\frac{m{v}^2}{R}=qvB,$ where, m = mass of the particle

$\Rightarrow R=\frac{mv}{qB}$

(g) Time taken to complete one revolution is

$T=\frac{2\pi R}{v}$ $\Rightarrow$ T = $\frac{2\pi m}{qB}$

Note: The time period is independent of the speed of the particle.

If the charged particle is projected neither parallel nor perpendicular to the field then its velocity $\vec{v}$can be resolved into two components, one along the $\overrightarrow{B}$ say v_|| and the other perpendicular to $\overrightarrow{B}$ say $V_{\perp }$_. It experiences a magnetic force and hence has a tendency to move on a circular path. Due to V_|| it experience no force, and hence has a tendency to move on a straight path along the field. So in this case it moves along a helical path.

(h) Radius of the helix is R = $\frac{m{v}_{\perp }}{qB}$

(i) Time taken to complete one revolution is $T=\frac{2\pi m}{qB}.$

(j) The distances moved by the charged particle along the magnetic field during one revolution is called pitch.

$\Rightarrow pitch=v_{II}\times T=\frac{2\pi m{v}_{II}}{qB}$

Illustration : An $\alpha$–particle is describing a circle of radius 0.45 m in a field of magnetic induction 1.2 weber/m². Find its speed, frequency of rotation and kinetic energy. What potential difference will be required which will accelerate the particle, so as to give this much energy to it? The mass of $\alpha$–particle is 6.8 × 10^–27 kg and its charge is 3.2 × 10^–27kg and its charge is 3.2 × 10^–19 coulomb.

Solution: We have

$F=qvB=\frac{m{v}^2}{r}$

or $v=\frac{qBr}{m}$ $=\frac{3.2\times {10}^{-19}\times 1.2\times 0.45}{6.8\times {10}^{-27}}$

= 2.6 x 10⁷ m/s

The frequency of rotation

$f=\frac{v}{2\pi r}$ $=\frac{2.6\times {10}^7}{2\times 3.14\times 0.45}$

= 9.2 x 10⁶ sec^–1.

Kinetic energy of $\alpha$–particle, $E_K=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 6.8\times {10}^{-27}\times {\left(2.6\times {10}^7\right)}^2$

= 2.3 × 10^–12 joule.

$=\frac{2.3\times {10}^{-12}}{1.6\times {10}^{-19}}eV=14\times {10}^{6}ev=14MeV.$

If V is accelerating potential of $\alpha$–particle, then Kinetic energy = qV

14 × 10⁶ eV = 2eV (since charge on $\alpha$–particle = 2e)

$\therefore$ $V=\frac{14\times {10}^6}{2}=7\times {10}^{6}Volt.$