Question

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B=B_0\hat{k}$.

• A. The charge to mass ratio satisfy:
  ${\left(\frac{e}{m}\right)}_1+{\left(\frac{e}{m}\right)}_2=0$.
• B. They necessarily represent a particle-antiparticle pair.
• C. They must have equal charges.
• D. They have equal $z$- components of momenta.

Answer 1

The correct option is A The charge to mass ratio satisfy:

${\left(\frac{e}{m}\right)}_1+{\left(\frac{e}{m}\right)}_2=0$.


As per the given diagram, if the particle is thrown in $x-y$ plane at some angle $\theta$ with velocity $v$, then by resolving the velocity of the particle in rectangular components, such that one component is along the field i.e., parallel to the field $\left(v\cos \theta \right)$ and other one is perpendicular to the field $\left(v\sin \theta \right)$.

Now according to diagram, particle moves with constant velocity $v\cos \theta$ along the field.

Then, the distance covered by the particle along the magnetic field is called pitch.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by

$p=T\left(v\cos \theta \right)$ ...$\left(i\right)$

Where $T$ is time period for one complete helical motion.

As we know, $T=\frac{2\pi m}{qB}$

By putting value of $T$ in $\left(i\right)$ equation we get,

$p=\frac{2\pi m}{qB}\left(v\cos \theta \right)$

For given pitch $p$ correspond to charge
particle, we get

$\frac{q}{m}=\frac{2\pi v\cos \theta }{pB}=$ constant

Therefore, charged particles transverse identical helical paths in a completely opposite sense in a uniform magnetic field $B.L.H.S$ for two particles should be same and of opposite sign.

Therefore, ${\left(\frac{e}{m}\right)}_1+{\left(\frac{e}{m}\right)}_2=0$

$\text{Final Answer}:$ (d)