Question 149 of 276

A particle of charge $q$ and mass $m$ is moving with a velocity $-v\hat{i}(v\neq 0)$ towards a large screen placed in the $Y-Z$ plane at distance d. If there is magnetic field $\vec{B}={B}_{0}\hat{k},$ the minimum value of $v$ for which the particle will not hit the screen is :

A$\dfrac{{qdB}_{0}}{3m}$B$\dfrac{2qd{B}_{0}}{m}$C$\dfrac{qd{B}_{0}}{m}$D$\dfrac{{qdB}_{0}}{2m}$

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Answer: (C) $\dfrac{qd{B}_{0}}{m}$

The radius of circular path in given magnetic field is
$r=d=\dfrac{mv}{q{B}_{0}}$
so $v=\dfrac{q{B}_{0}d}{m}$