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4.Moving Charges and Magnetism
medium
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 a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is
A
$\frac{ q d B _{0}}{2 m }$
B
$\frac{q d B_{0}}{m}$
C
$\frac{2 q d B_{0}}{m}$
D
$\frac{q d B_{0}}{3 m}$
(JEE MAIN-2020)
Solution

In uniform magnetic field particle moves in a circular path, if the radius of the circular path is $'d',$ particle will not hit the screen.
$d =\frac{ mv }{ qB _{0}}$
$v =\frac{ q B _{0} d }{ m }$
$\therefore$ correct option is $(2)$
Standard 12
Physics