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Q.

A particle of mass m, carrying a charge Q is accelerated through a potential difference V and then allowed to pass through a uniform magnetic field B in a direction normal to the field. The radius of its trajectory is found to be R. If charge of the particle is halved and accelerating potential is made four times its initial value, then radius of the trajectory will be

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a

$2 \sqrt{2} R$

b

$\frac{R}{\sqrt{2}}$

c

$\frac{R}{2}$

d

$\sqrt{2} R$

answer is D.

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Detailed Solution

$\frac{1}{2} {mu}^{2} = Q V \Rightarrow u = \sqrt{\frac{2 Q V}{m}}$

$∴ R = \frac{u}{ω} = \frac{\sqrt{2 QV / m}}{QB / m} = \frac{1}{B} \sqrt{\frac{2 Vm}{Q}}$

$∴ \frac{R'}{R} = \sqrt{\frac{4 V}{V}} . \sqrt{\frac{Q}{Q / 2}} = 2 \sqrt{2} \Rightarrow R' = 2 \sqrt{2} R$

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A particle of mass m, carrying a charge Q is accelerated through a potential difference V and then allowed to pass through a uniform magnetic field B in a direction normal to the field. The radius of its trajectory is found to be R. If charge of the particle is halved and accelerating potential is made four times its initial value, then radius of the trajectory will be