A charge particle having charge q is accelerated through a potential difference V and then it enters a perpendicular magnetic field in which it experiences a force F. If V is increased to 5V, the particle will experience a force
A
F
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B
5F
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C
F5
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D
√5F
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Solution
The correct option is D√5F 12mv2=qV⇒v=√2qVm. Also F=qvB ⇒F=qB√2qVm hence F∝√V which gives F′=√5F.