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When an electron is accelerated through a p.d 'V', it experiences a force F through a uniform transverse magnetic field. If the P.D is increased to 2V, the force experienced by the electron in the same magnetic field is

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a

$2 F$

b

$2 \sqrt{2} F$

c

$\sqrt{2} F$

d

$4 F$

answer is C.

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Detailed Solution

$P.D = V_{1} = V Force = F_{1} = F Magneticfield = B P.D = V_{2} = 2V F_{2} = ? (B is same) F = Bqv = Bq \sqrt{\frac{2Vq}{m}} = \sqrt{\frac{B^{2} q^{3} 2V}{m}} F \propto \sqrt{V} \frac{F_{1}}{F_{2}} = \sqrt{\frac{V_{1}}{V_{2}}} \frac{F}{F_{2}} = \sqrt{\frac{V}{2V}} \frac{F}{F_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} F_{2} = \sqrt{2} F$

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