A) \[{{10}^{-8}}\,m\]
B) \[2\times {{10}^{-8}}\,m\]
C) \[{{10}^{-6}}\,m\]
D) \[{{10}^{-10}}\,m\]
Correct Answer: A
Solution :
If electron moves in a magnetic field at an angle \[\theta \] (other than \[{{0}^{\text{o}}},\,{{180}^{\text{o}}}\] or \[{{90}^{\text{o}}}\]), its velocity can be resolved in two components one along \[\vec{B}\] and another perpendicular to \[\vec{B}\]. Let the two components be \[{{v}_{||}}\] and \[{{v}_{\bot }}\] Then \[{{v}_{||}}=v\cos \theta \] \[and{{v}_{\bot }}=v\sin \theta \] The component perpendicular to field (\[{{v}_{\bot }}\]) gives a circular path and the component parallel to field (\[{{V}_{||}}\]) gives a straight line path. The resultant path is, helix as shown in figure. The radius of this helical path is \[r=\frac{m{{v}_{\bot }}}{eB}\] \[=\frac{mv\sin \theta }{eB}\] \[orr=\frac{v\sin \theta }{\left( \frac{e}{m} \right)B}\] Given \[v=1\times {{10}^{3}}\,m/s,\,B=0.3\,T,\,\theta ={{30}^{o}},\] \[\frac{e}{m}=1.76\times {{10}^{11}}\,C/kg\] \[\therefore r=\frac{1\times {{10}^{3}}\sin {{30}^{o}}}{1.76\times {{10}^{11}}\times 0.3}\] \[=\frac{1\times {{10}^{3}}\times \frac{1}{2}}{1.76\times {{10}^{11}}\times 0.3}={{10}^{-8}}\,m\]You need to login to perform this action.
You will be redirected in
3 sec