question_answer

An electron moves with a velocity \[1\times {{10}^{3}}\,m/s\] in a magnetic field of induction \[0.3\text{ }T\] 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:        [AIPMT 2000]

A)  \[{{10}^{-8}}\,m\]     

B)       \[2\times {{10}^{-8}}\,m\]

C) \[{{10}^{-6}}\,m\]

D)                   \[{{10}^{-10}}\,m\]

Correct Answer: A

Solution :

[a] If electron moves in a magnetic field at an angle \[\theta \] (other than \[0{}^\circ ,\text{ }180{}^\circ \] or \[90{}^\circ \]), 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

\[r=\frac{m{{v}_{\bot }}}{eB}\]

\[=\frac{mv\sin \theta }{eB}\]

or         \[r=\frac{v\sin \theta }{\left( \frac{e}{m} \right)B}\]

Given,   \[v=1\times {{10}^{3}}\,m/s,\,B=0.3\,T,\,\theta ={{30}^{o}},\]

\[\frac{e}{m}=1.76\times {{10}^{11}}\,C/kg\]

\[\therefore \]      \[r=\frac{1\times {{10}^{3}}\sin {{30}^{o}}}{1.76\times {{10}^{11}}\times 0.3}\]

\[=\frac{1\times {{10}^{3}}\times \frac{1}{2}}{1.76\times {{10}^{11}}\times 0.3}={{10}^{-8}}\,m\]