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Question: Answered & Verified by Expert
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $\mathbf{B}=\mathrm{B}_0 \hat{\mathbf{k}}$.
PhysicsMagnetic Effects of Current
Options:
  • A
    They have equal z-components of momenta
  • B
    They must have equal charges
  • C
    They necessarily represent a particle, anti-particle pair
  • D
    The charge to mass ratio satisfy
    $$
    \left(\frac{\mathrm{e}}{\mathrm{m}}\right)_1+\left(\frac{\mathrm{e}}{\mathrm{m}}\right)_2=0
    $$
Solution:
1880 Upvotes Verified Answer
The correct answer is:
The charge to mass ratio satisfy
$$
\left(\frac{\mathrm{e}}{\mathrm{m}}\right)_1+\left(\frac{\mathrm{e}}{\mathrm{m}}\right)_2=0
$$
As we know that the uniqueness of helical path is determined by its pitch
$$
\mathrm{P}(\text { Pitch })=\frac{2 \pi \mathrm{mv} \cos \theta}{\mathrm{Bq}}
$$
Where $\theta$ is angle of velocity of charge particle with $x$-axis For the given pitch d correspond to charge particle, we have
$$
\frac{\mathrm{q}}{\mathrm{m}}=\frac{2 \pi \mathrm{v} \cos \theta}{\mathrm{BP}}=\mathrm{constant}
$$
If motion is not helical, $(\theta=0)$
As charged particles traverse identical helical paths in a completely opposite direction in a same magnetic field $\mathbf{B}$, LHS for two particles should be same and of opposite sign.
$$
\therefore\left(\frac{\mathrm{e}}{\mathrm{m}}\right)_1+\left(\frac{\mathrm{e}}{\mathrm{m}}\right)_2=0
$$

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