Question

Two particles, each having a mass $m$ are placed at a separation $d$ in a uniform magnetic field $B$ as shown in figure. They have opposite charge and they are projected towards each other, each with a speed $v$. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value $v_m$ of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if $v=v_m/2$? (c) At what instant will a collision occur between the particles if $v=2v_m$? (d) Suppose $v=2v_m$ and the collision between the particles is completely inelastic. Describe the motion after the collision.

Answer 1

$\bullet$ For a charged particle (mass(m),charge(q)) with a velocity $V$ perpendicular to a uniform

magnetic field $B$, it moves in a circular path with radius $r=\frac{mv}{qB}$

a. Refer Fig. 1

For mas v, closest approach should be zero, hence the $r=\frac{d}{2}$

So, $v_m=\frac{dqB}{2m}$

b. Refer Fig.2

From diagram, for $v=v_{m/2}$, the min and max separation,

$S_{max}=d+\frac{2mv}{qB}$ and $s_{min}=d-\frac{2mv}{qB}$

$S_{max}=d+\frac{m{v}_m}{qB}$ and $s_{min}=d-\frac{m{v}_m}{qB}$

c. Refer Fig. 3

from the diagram, $\theta ={\sin }^{-1}\left(\frac{d}{2r}\right)$

as, $v=2v_m$

$W=\frac{v}{r}=\frac{2\pi }{T}$, where $r=\frac{mv}{qB}$ and $T$ is time period of rotation

So, collision happens after time $t$

$t=\frac{\theta }{2\pi }T=\frac{m}{qB}{\sin }^{-1}\left(\frac{1}{2qB}\right)seconds$

d. As the collision is completely inelastic, after the two particles will act as a particle with no charge

(hence no interaction with magnetic field) which moves in a straight line with a speed, $v=2v_m$