Directions : (16-20) |
Motion of Charge in Magnetic Fields |
An electron with speed \[{{v}_{0}}<<c\] moves in a circle of radius \[{{r}_{0}}\] in a uniform magnetic field. This electron is able to traverse a circular path as magnetic field is perpendicular to the velocity of the electron. A force acts on the particle perpendicular to both \[{{\overrightarrow{v}}_{0}}\] and \[{{\overrightarrow{B}}_{0}}\]. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is \[{{T}_{0}}\]. |
A) \[4{{r}_{0}}\]
B) \[2{{r}_{0}}\]
C) \[{{r}_{0}}\]
D) \[{{r}_{0}}/2\]
Correct Answer: B
Solution :
As, \[{{r}_{0}}=\frac{mv}{qB}\Rightarrow r'=\frac{m\left( 2{{v}_{0}} \right)}{qB}=2{{r}_{0}}\]You need to login to perform this action.
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