Focusing a beam of ions

If any charged particle, say an electron is released in magnetic field with velocity at an angle $\alpha$ , it will undergo helical motion because of:

perpendicular component of velocity that makes electron undergo circular motion and parallel component of velocity along the magnetic field that remains unaffected.

Hence, $\frac {mv^2_{perp}}{r} = qv_{perp} B$

where, r = radius of circle q = charge of charged particle B = magnetic field strength m = mass of charged particle

therefore, $v_{perp} = \frac {qbr}{m}$

and T ,time period = $\frac {2\pi r}{v_{perp}} = \frac {2\pi m}{qB}$

So, $pitch = \overline{OA}$ = distance moved in forward direction along the magnetic field

$\Rightarrow \overline{OA} = v_{parallel} \times T$

$= v \cos\alpha \times \frac {2\pi m}{qB}$

$v \cos\alpha = v$ because $\alpha \ll 1$ , assuming $\alpha \approx 0$

$= v \times 2\pi \frac {m}{q} \times \frac{1}{B}$

$=100 \times 10^{-5} \cdot 10^{3} \times 2\pi$

$=2\pi$

Therefore, $\overline{OA} = 2\pi$ or 6.28 meters

First, consider the motion along the $z$ -axis. The velocity of the particle along the $z$ -axis is $v \cos{\alpha_0}$ , so the distance $OA = t\cdot v\cos{\alpha_0}$ (let's call it $(*)$ ), where $t$ is the time the particle needs to reach point $A$ .

Next, consider the motion of the particle that is perpendicular to $z$ -axis. In this case, the particle moves with velocity $v\sin{\alpha_0}$ . Since the particle is charged, it experiences the Lorentz force that is perpendicular to the velocity (the direction of the force is irrelevant in our case) given by $F_L = qvB\sin{\alpha_0}$ .

This force makes the particle follow a circular path and imparts centripetal acceleration $a = \frac{v^2}{r}$ . Using Newton's Second Law and the expression for the Lorentz force, we can derive the expression for the radius to be $r = \frac{v\sin{\alpha_0}}{\frac{q}{m}B}$ .

The time it takes for a particle to complete a loop is $t = \frac{2\pi r}{v\sin{\alpha_0}}$ . In our case, it is the same time that is required for a particle to reach point $A$ along the $z$ -axis. Therefore, we can substitute this expression for $t$ into $(*)$ to find $OA$ .

$$OA = \frac{v\cos{\alpha 0} 2\pi r}{v \sin{\alpha 0}} = \cot{\alpha 0} * 2\pi r = \cot{\alpha 0} * 2\pi \frac{v \sin{\alpha 0}}{\frac{q}{m} B} = 2 \pi \cos{\alpha 0} \frac{v}{\frac{q}{m} B}$$

Since $\alpha_0 \ll 1$ , we can use the small angle approximation and consider $\cos{\alpha_0} = 1$ . Hence,

$$OA = \frac{2 \pi v}{\frac{q}{m} B} = \frac{2 * 3.14 * 100}{10^5 \cdot 10^{-3}} = 6.28 (m)$$

$A$ = pitch = $vcos\alpha_{0} T$ = $v \frac{2 \pi m}{qB}$ = $\fbox{6.28}$ , as $\alpha_{0} << 1$

We can divide this velocity vector into two components, v cos(alpha) and v sin(alpha). Since F = q (v x B) , the horizontal component i.e. v cos(alpha) will be unaffected by B, Only v sin(alpha) will be affected . The magnetic field will keep on changing the direction of v sin(alpha) and this component's locus will be circle, where q v sin(alpha) B = m v^2 sin^2 (alpha) /r

or r /v sin(alpha) = m/qB or 2pi r/v sin(alpha) = 2pi m/qB note that 2pi r/ v sin(alpha) is the time taken by particle to complete one revolution i.e. reach point A. Hence distance OA = v cos(alpha) * 2pi m/qB since alpha is very small, we can assume cos(alpha) = 1 , and now substituting the respective values of m/q,B and v, we get our desired answer i.e. 2pi