Two particle A and B of masses m A and m B respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles a

Question

Two particle A and B of masses m A ​ and m B ​ respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are v A ​ and v B ​ respectively and the trajectories are as shown in the figure. Then-

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Solution:

When a charged particle moves perpendicular to a uniform magnetic field, it undergoes circular motion. The magnetic force provides the centripetal force:

$F_{B} = q v B = \frac{m v ^{2}}{r}$

From this, the radius of the circular path is $r = \frac{m v}{qB}$ .

Since the charge $q$ and magnetic field $B$ are the same for both particles, the radius $r$ is directly proportional to the momentum $m v$ .

From the figure, the radius of particle A's trajectory ( $r_{A}$ ) is greater than the radius of particle B's trajectory ( $r_{B}$ ).

Therefore, $r_{A} > r_{B}$ , which implies $m_{A} v_{A} > m_{B} v_{B}$ .

The final answer is $B$

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