A) 0.2cm
B) 1.2cm
C) 2.2cm
D) 0.122cm
Correct Answer: B
Solution :
When the charged particle is moving at an angle to the field (other than \[{{0}^{o}},{{90}^{o}}\] and \[{{180}^{o}}\]), in this situation resolving the velocity of the particle along and perpendicular to the field, we find that the particle moves with constant velocity v \[\cos \theta \] along the field and at the same time it is also moving with velocity v sin \[\theta \] perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field) of radius \[r=\frac{m(v\sin \theta )}{qB}\] Here, \[m=1.67\times {{10}^{-27}}kg\] \[v=4\times {{10}^{5}}m/s\] \[\theta ={{60}^{o}}\] \[q=1.6\times {{10}^{-19}}C\] \[B=0.3\,T\] \[\therefore \] \[r=\frac{1.67\times {{10}^{-27}}\times 4\times {{10}^{5}}\times (\sqrt{3}/2)}{1.6\times {{10}^{-19}}\times 0.3}\] \[=0.012\,m\] \[=1.2\,\,cm\]
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