4.3 MOTION IN A MAGNETIC FIELD
We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 6) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, \( q\mathbf{E} \), which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.]
We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of \( \mathbf{v} \) perpendicular to \( \mathbf{B} \). The perpendicular force, \( q\mathbf{v × B} \), acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if \( \mathbf{v} \) and \( \mathbf{B} \) are perpendicular to each other (Fig. 4.5).
If velocity has a component along \( \mathbf{B} \), this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to \( \mathbf{B} \) is as before a circular one, thereby producing a helical motion (Fig. 4.6).
You have already learnt in earlier classes (See Class XI, Chapter 4) that if \( r \) is the radius of the circular path of a particle, then a force of \( m v^2 / r \), acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the velocity \( \mathbf{v} \) is perpendicular to the magnetic field \( \mathbf{B} \), the magnetic force is perpendicular to both \( \mathbf{v} \) and \( \mathbf{B} \) and acts like a centripetal force. It has a magnitude \( q v B \). Equating the two expressions for centripetal force, \[ m v^{2} / r=q v B \] which gives
\[ r=mv/qB \](4.5)
for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If \( \omega \) is the angular frequency, then \( v=\omega r \). So,
\[ \omega=2 \pi v=q B / m \](4.6(a))
which is independent of the velocity or energy . Here \( v \) is the frequency of rotation. The independence of \( v \) from energy has important application in the design of a cyclotron (see Section 4.4.2).
The time taken for one revolution is \(T=2 \pi / \omega \equiv 1 / v\). If there is a component of the velocity parallel to the magnetic field (denoted by \( v_{||} \)), it will make the particle move along the field and the path of the particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch \( p \). Using Eq. [4.6 (a)], we have
\[ p=v_{||} T=2 \pi m v_{||} / q B \](4.6(b))
The radius of the circular component of motion is called the radius of the helix.
Example 4.3
What is the radius of the path of an electron (mass \( 9 × 10^{-31} kg \) and charge \( 1.6 × 10^{–19} C \)) moving at a speed of \( 3 ×10^7 m/s \) in a magnetic field of \( 6 × 10^{–4} T \) perpendicular to it? What is its frequency? Calculate its energy in \( keV \). (\( 1 eV = 1.6 × 10^{–19} J \)).
Using Eq. (4.5) we find \[ \begin{array}{1} &r = m v /(q B) \\ &=\frac{9 \times 10^{-31} \mathrm{~kg} \times 3 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}}{(1.6 \times 10^{-19} \mathrm{C} \times 6 \times 10^{-4} \mathrm{~T})} \\ &=26 \times 10^{-2}m=26 cm \end{array} \] \[ \begin{array} &frequency\text{ v} =&v /(2 \pi r)=2 \times 10^{6} \mathrm{~s}^{-1} \\ &=2 \times 10^{6} \mathrm{~Hz}=2 \mathrm{MHz} \end{array}\] \[ \begin{array} &E=(1 / 2) m w^{2}\\ =(1 / 2) 9 \times 10^{-31} \mathrm{~kg} \times 9 \times 10^{14} \mathrm{~m}^{2} / \mathrm{s}^{2}\\ =40.5 \times 10^{-17} \mathrm{~J} \\ \approx 4 \times 10^{-16} \mathrm{~J} =2.5 \mathrm{keV} \end{array} \]
Helical Motion of Charged Particles and Aurora Borealis
In polar regions like Alaska and Northern Canada, a splendid display of colours is seen in the sky. The appearance of dancing green pink lights is fascinating, and equally puzzling. An explanation of this natural phenomenon is now found in physics, in terms of what we have studied here.
Consider a charged particle of mass m and charge q, entering a region of magnetic field \( \mathbf{B} \) with an initial velocity \( \mathbf{v} \). Let this velocity have a component \( \mathbf{v}_p \) parallel to the magnetic field and a component \( \mathbf{v}_n \) normal to it. There is no force on a charged particle in the direction of the field. Hence the particle continues to travel with the velocity \( \mathbf{v}_p \) parallel to the field. The normal component \( \mathbf{v}_n \) of the particle results in a Lorentz force (\( \mathbf{V}_n \mathbf{×B} \)) which is perpendicular to both \( \mathbf{v}_n \) and \( \mathbf{B} \). As seen in Section 4.3.1 the particle thus has a tendency to perform a circular motion in a plane perpendicular to the magnetic field. When this is coupled with the velocity parallel to the field, the resulting trajectory will be a helix along the magnetic field line, as shown in Figure (a) here. Even if the field line bends, the helically moving particle is trapped and guided to move around the field line. Since the Lorentz force is normal to the velocity of each point, the field does no work on the particle and the magnitude of velocity remains the same.
During a solar flare, a large number of electrons and protons are ejected from the sun. Some of them get trapped in the earth’s magnetic field and move in helical paths along the field lines. The field lines come closer to each other near the magnetic poles; see figure (b). Hence the density of charges increases near the poles. These particles collide with atoms and molecules of the atmosphere. Excited oxygen atoms emit green light and excited nitrogen atoms emits pink light. This phenomenon is called Aurora Borealis in physics.