question_answer

                An electron moves with a velocity \[1\times {{10}^{3}}\,m/s\] in a magnetic field of induction 0.3 T at an angle \[{{30}^{o}}\]. If \[\frac{e}{m}\] of electron is \[1.76\times {{10}^{11}}\,C/kg\] the radius of the path is nearly:                                                 

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 \[v=1\times {{10}^{3}}\,m/s,\,B=0.3\,T,\,\theta ={{30}^{o}},\]                 \[\frac{e}{m}=1.76\times {{10}^{11}}\,C/kg\]                 \[\therefore r=\frac{1\times {{10}^{3}}\sin {{30}^{o}}}{1.76\times {{10}^{11}}\times 0.3}\] \[=\frac{1\times {{10}^{3}}\times \frac{1}{2}}{1.76\times {{10}^{11}}\times 0.3}={{10}^{-8}}\,m\]