A magnetic field of 0.3T is present in a region and a proton enters the region with a velocity 4 x 105 m/s making an angle of 600 with the field. If the proton completes 10 revolutions by the time it crosses the region as shown in the figure, then is close to (mass of proton=1.67 x 10-27 kg, a charge of proton=1.6 x 10-19 C, Answer in meter )
Option: 1 0.44
Option: 2 0.88
Option: 3 0.22
Option: 4 0.11
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Given
m = 1.67 x 10-27 kg
v = 4 x 105 m/s
q = 1.6 x 10-19 C
B = 0.3 T
Using
we get