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Two particles 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, Then

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m_{A} v_{A}

m_{B} v_{B}

m_{A} v_{A}

m_{B} v_{B}

m_{A}

m_{B}

and

v_{A}

v_{B}

m_{A}

m_{B}

and

v_{A}

v_{B}

answer is B.

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Detailed Solution

The speeds of the particles are

v_{A}

and

v_{B}

respectively and the trajectories are as shown in the, Then

m_{A} v_{A} > m_{B} v_{B}

As per the formula

r = \frac{p}{qB}

where,

r = radius

p = particle

q = charge of particle

B = external magnetic field

when a particle charged is moving at right angle to the magnetic field, then a force acts on it which behaves as a centripetal force and moves the particle in circular path.

r_{B} < r_{A} p_{B} < p_{A}

m_{B} v_{B} < m_{A} v_{A}

m_{A} v_{A} > m_{B} v_{B}

Where,

m = masses

v = speed of particle

p = particle

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