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Circular Kinematics

A charged par...

Question

A charged particle of mass m and charge q is accelerated through a potential difference of V volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by

\frac{V m}{q B^{2}}

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\frac{2 V m}{q B^{2}}

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\sqrt{\frac{2 V m}{q}} . (\frac{1}{B})

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\sqrt{\frac{V m}{q}} . (\frac{1}{B})

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Solution

The correct option is C $\sqrt{\frac{2 V m}{q}} . (\frac{1}{B})$
The speed of particle just before entering into magnetic field is $v_{0}$ .
$∴ q V = \frac{1}{2} m v_{0}^{2}$
$\Rightarrow v_{0} = \sqrt{(\frac{2 q V}{m})}$
Radius of the circular path is given by $r = \frac{m v_{0}}{q B}$
$∴ r = \frac{m}{q B} \sqrt{(\frac{2 q V}{m})} = \frac{1}{B} \sqrt{(\frac{2 m V}{q})}$

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