Charged particle moving in a uniform magnetic field
Case I. When motion of charged particle is parallel or antiparallel to the uniform magnetic field.
In this case, the velocity $\vec{v}$ of the charged particle makes an angle $0^\circ$ or $180^\circ$ with the direction of the magnetic field. Therefore, magnetic force acting on the charged particle is given by
$\vec{F}_m = q (\vec{v} \times \vec{B}) = qvB \sin \theta$
Since $\theta = 0^\circ$ or $180^\circ$,
so $\sin 0^\circ = 0$ or $\sin 180^\circ = 0$
$\vec{F}_m = 0$
Thus, moving charged particle does not experience any force. Therefore, the charged particle continues to move along a straight line if it moves parallel or anti-parallel to the direction of the uniform magnetic field.
Case II. When charged particle moves at right angle to the magnetic field.
Consider a uniform magnetic field ($\vec{B}$) acting perpendicular to the plane of the paper and directed into the paper. Let a charged particle having charge $+q$ enter the region of the magnetic field with velocity $\vec{v}$ and perpendicular to the direction of magnetic field $\vec{B}$.
Magnetic force acting on the charge $+q$ due to the magnetic field is given by,
$\vec{F} = q (\vec{v} \times \vec{B}) = qvB \sin \theta$
or
$(\because \theta = 90^\circ)$
$F = qvB \sin 90^\circ$
or
$F = qvB \quad \dots (i)$
This force is always perpendicular to the direction of motion of the particle as well as the magnetic field. Thus, the path of the charged particle is circular.
Radius of circular path :
If $m$ be the mass of the charged particle and $r$ be the radius of the circular path, then the necessary centripetal force required by the particle to move in a circular path is given by
$F_c = \frac{mv^2}{r}$.
This centripetal force is provided by the magnetic force, $F = qvB$.
$\frac{mv^2}{r} = qvB$
or
$r = \frac{mv}{qB}$
$r = \frac{mv}{qB} = \frac{v}{(q/m)B} \quad \dots (ii)$
Thus, radius of the circular path of a charged particle in a uniform magnetic field is
[(i)] directly proportional to the magnitude of the momentum ($mv$) of the particle.
[(ii)] directly proportional to the mass ($m$) of the particle.
[(iii)] directly proportional to the speed ($v$) of the particle.
[(iv)] inversely proportional to the magnitude of the charge on the particle.
[(v)] inversely proportional to the magnitude of the magnetic field.
Kinetic energy of the charged particle
$K.E= \frac{1}{2} mv^2$
$( \because v = \frac{qBr}{m})$
$K.E= \frac{1}{2} m ( \frac{qBr}{m})^2$
$K.E= \frac{1}{2}.\frac{q^{2}B^{2}r^{2}}{2m^{2}}$
Time period of the charged particle to complete a circular path of radius $r$ is given by
$T = \frac{\text{Distance}}{\text{Speed}}$
$T= \frac{\text{Circumference of the circle}}{Speed}$
$T= \frac{2\pi r}{v}$
Using eqn. (ii), we have
$r = \frac{mv}{qB}
Or
$v = \frac{qBr}{m}$
$T = \frac{2\pi r}{v}$
$T = \frac{2\pi r}{qBr/m}$
$T = \frac{2\pi m}{qB} \quad \dots (iv)$
Frequency of the charged particle :
$\nu = \frac{1}{T} = \frac{qB}{2\pi m} \quad \dots (v)$
From equation (iv) and (v), it is clear that time period (or frequency) is independent of speed of particle and radius of the circular path. It depends only on the magnetic field $B$ and the nature of the particle, i.e., specific charge $q/m$. Thus, all the charged particles take the same time to complete the circular orbits of small or large radius, provided their specific charge ($q/m$) is same.
This fact is used to accelerate charged particles in a Cyclotron. This principle is also used in mass-spectrometer* or bubble-chamber (a particle detector).
Angular frequency:
$\omega = 2\pi \nu = \frac{qB}{m}$
or
$\omega = \frac{v}{r} = \frac{qBr/m}{r} = \frac{qB}{m} \quad \dots (vi)$
This angular frequency is called \textbf{gyro frequency}.
From equation (vi), it is clear that angular frequency does not depend upon the speed of the particle (provided speed of the particle is very small as compared to the speed of the light).
Case III. When the charged particle moves at an angle to the magnetic field (other than $0^\circ$, $90^\circ$ and $180^\circ$).
Consider a charged particle of mass $m$ moving with velocity $\vec{v}$ at an angle $\theta$ with the direction of magnetic field $\vec{B}$ along x-axis (in figure 13).
Now $v_x = v \cos \theta$ is the component of the velocity $\vec{v}$ along the direction of $\vec{B}$ and $v_y = v \sin \theta$ is the component of $\vec{v}$ perpendicular to the direction of $\vec{B}$.
The charged particle moves with constant velocity $v \cos \theta$ along the direction of the magnetic field, so no magnetic force acts on the charged particle in the direction of $\vec{B}$ ($F = q (v \cos \theta) B \sin 0^\circ = 0$).
Since $v \sin \theta$ is perpendicular to the direction of $\vec{B}$, so the particle experiences a magnetic force ($F = q (v \sin \theta) B$) perpendicular to the magnetic field. Under this magnetic force charged particle tends to move in a circular path in a plane perpendicular to the magnetic field.
Centripetal force required to move the particle in a circle is provided by the magnetic force.
$\frac{m (v \sin \theta)^2}{r} = q (v \sin \theta) B$
or
$r = \frac{m (v \sin \theta)^{2}}{qBv \sin \theta} $
$r=\frac{m v^{2}\sin \theta}{qvB}$
Thus, radius of the circular path $r = \frac{m v \sin \theta}{qB} \quad \dots (i)$
Time period of the particle.
$T = \frac{\text{Distance}}{\text{Speed}} = \frac{2\pi r}{v \sin \theta}$
$T = \frac{2\pi}{v \sin \theta} \times \frac{m v \sin \theta}{qB} = \frac{2\pi m}{qB} \quad \dots (ii)$
and frequency,
$\nu = \frac{1}{T} = \frac{qB}{2\pi m}$
or
$\omega = 2\pi \nu = \frac{qB}{m} \quad \dots (iii)$
Shape of path of the charged particle :
The charged particle is under the combined effect of two components of velocity. It moves along the direction of the magnetic field due to the horizontal component ($v \cos \theta$) of the velocity and at the same time moves in a circular path due to ($v \sin \theta$) component of the velocity. Thus the resultant path of the particle is a helix with its axis parallel to the magnetic field $\vec{B}$.
Pitch of helix:
The linear distance travelled by the charged particle along the direction of magnetic field in one rotation is called \textbf{pitch} of the helix.
$p = (v \cos \theta) \times T$
$p = (v \cos \theta) \times \frac{2\pi m}{qB}$ [Using eqn. (ii)]
or
$p = \frac{2\pi m v \cos \theta}{qB}$
or
$p = \frac{2\pi (mv) \cos \theta}{qB}$