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Question

A cyclotron is used to accelerate protons to a kinetic energy of 5 MeV. If the strength of the applied magnetic field in the cyclotron is 2 T, the radius of the cyclotron will be
(Velocity of proton is 3×107 m/s)

A
0.241 m
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B
0.167 m
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C
0.423 m
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D
0.765 m
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Solution

The correct option is B 0.167 m
Given,
The kinetic energy (K) of proton,

K=5 MeV=5×106×(1.6×1019) J

K=8.0×1013 J

The magnetic force acting on proton will act as source of centripetal force, so

qvB=mv2rqvB2=12mv2r=Kr

r=2KqvB

Substituting the values, we get

r=2×8.0×10131.6×1019×3×107×2=106×101

r=0.167 m

Hence, option (b) is correct answer.

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