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Q.

A charged particle, having charge $q_{1}$ accelerated through a potential difference $V$ enters a perpendicular magnetic field in which it experiences a force F. If V is increases to $5 V$ . The particle will experience a force

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a

$F / 5$

b

$\sqrt{5} F$

c

5F

d

F

answer is D.

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Detailed Solution

$K E = \frac{1}{2} m v^{2} = q V v = \sqrt{\frac{2 g}{M}} = \sqrt{v} F \propto \sqrt{v} \frac{F'}{F} = \sqrt{\frac{5 v}{u}} = 5 \times 10^{- 4}$

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