Question

A slightly divergent beam of charged particles accelerated by a potential difference V
 propagates from a point A along the axis of solenoid. The beam is brought into focus at a distance l from the point A at two successive values of magnetic induction ${{B}_{1}}$ and ${{B}_{2}}$. If the specific charge $\dfrac{q}{m}$ of the particles is $\dfrac{q}{m}=\dfrac{x{{\pi }^{2}}V}{{{l}^{2}}{{\left( {{B}_{2}}-{{B}_{1}} \right)}^{2}}}$. Find x.

Answer 1

Hint: Use the work energy theorem and find the velocity of the charged particles. The particles will move in a helical path with time period $T=\dfrac{2\pi m}{qB}$. Use the formula for pitch i.e. $p={{v}_{||}}\times T$. Find the value of p for consecutive focusing and with these two equations find $\dfrac{q}{m}$.

Formula used:
W = qV
$W=\Delta K$
$K=\dfrac{1}{2}m{{v}^{2}}$
$T=\dfrac{2\pi m}{qB}$
$p={{v}_{||}}\times T$
$l=np$

Complete step by step answer:
It is said that a beam of charged particles are accelerated by a potential difference V. Due the potential difference, the electric field will do some work on a charged particle which will be equal to W = qV, where q is the charge on the particle.
According to the work energy theorem, work done on a particle is equal to the change in its kinetic energy. i.e. $W=\Delta K$.
In this case, let us assume the particles were at rest initially and after crossing the potential difference, they attain a velocity v. Let the mass of each particle be m. Hence, the change in kinetic energy is $\Delta K=\dfrac{1}{2}m{{v}^{2}}-0$
Hence, we get that
$qV=\dfrac{1}{2}m{{v}^{2}}$.
$\Rightarrow v=\sqrt{\dfrac{2qV}{m}}$ …. (i)
This means that the particles enter a magnetic field with velocity $v=\sqrt{\dfrac{2qV}{m}}$.
However, the charged particles are slightly diverged, they will make a small angle $\theta $ with the magnetic field. Hence, they will follow a helical path.
Since the angle is very small, $\cos \theta \approx 1$.
The time period of the motion of the particle is given as $T=\dfrac{2\pi m}{qB}$.
When the particle is under helical path, we define its pitch (p). It is the distance travelled along the direction of magnetic field in time that it takes to complete one rotation (i.e. T).
Hence, $p={{v}_{||}}\times T$.
Here, ${{v}_{||}}$ is the velocity of the particle along the magnetic field. And ${{v}_{||}}=v\cos \theta \approx v$.
Therefore,
$p=v\times T$ …. (ii).
Substitute the values of v and T in equation (ii).
$p=\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{qB}$
Particles are focused if l contains integral multiples of the pitch.
 $\Rightarrow l=np$ , (n=1,2,3…).
$\Rightarrow p=\dfrac{l}{n}$
$\Rightarrow p=l,\dfrac{l}{2},\dfrac{l}{3}.....$
For two consecutive focusing,
$l=\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{q{{B}_{1}}}$ and $\dfrac{l}{2}=\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{q{{B}_{2}}}$
This implies that
${{B}_{1}}=\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{ql}$ and ${{B}_{2}}=\sqrt{\dfrac{2qV}{m}}\times \dfrac{4\pi m}{ql}$
${{B}_{2}}-{{B}_{1}}=\sqrt{\dfrac{2qV}{m}}\times \dfrac{4\pi m}{ql}-\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{ql}$
${{B}_{2}}-{{B}_{1}}=\sqrt{\dfrac{2qV}{m}}\times \dfrac{2\pi m}{ql}$
$\Rightarrow {{B}_{2}}-{{B}_{1}}=\sqrt{\dfrac{2Vm}{q}}\times \dfrac{2\pi }{l}$
Square both the sides.
$\Rightarrow {{\left( {{B}_{2}}-{{B}_{1}} \right)}^{2}}=\dfrac{2Vm}{q}\times \dfrac{4{{\pi }^{2}}}{{{l}^{2}}}$
$\Rightarrow \dfrac{q}{m}=\dfrac{8{{\pi }^{2}}V}{{{l}^{2}}{{\left( {{B}_{2}}-{{B}_{1}} \right)}^{2}}}$
Hence, x=8.

Note: Note that if motion of the charges were not diverging and were along the direction of the magnetic field, then they would travel in a straight line without being affected by the magnetic field. This because the angle between the velocity of the particle and the magnetic field will be zero and the magnitude of magnetic field will be $F=qvB\sin \theta =qvB\sin 0=0$