A proton (mass =1.67×10−27kg and charge =1.6×10−19C is projected in a uniform magnetic field of 2 Wb/m2 with a velocity 3.4×107m/s in a direction perpendicular to the field. The acceleration of the proton should be
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a
6.5×1015m/s2
b
6.5×1013m/s2
c
6.5×1011m/s2
d
6.5×109m/s2
answer is A.
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Detailed Solution
F = ma = qvB⇒a =qvBm=1.6×10−19×2×3.4×1071.67×10−27=6.5×1015 m/s2