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JEE Main 10 April 2015 Paper

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Question : 19

Total: 90

A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width 'd'. If 'α' be the angle of deviation of proton from initial direction of motion (see figure), the value of sin a will be :

B
2
√
qd
mV
B
d
√
q
2mV
Bd√
q
2mV
qV√
Bd
2m

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