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-Zplane at a distance $d$. If there is a magnetic field $\vec{B}=B_{0} \hat{k}$, the minimum value of $v$ for which the particle will not hit the screen is:
Correct Option: , 3
(3) In uniform magnetic field particle moves in a circular path, if the radius of the circular path is ' $r$ ', particle will not hit the screen.
$r=\frac{m v}{q B_{0}}$ $\left[\because \frac{m v^{2}}{r}=q v B_{0}\right]$
Hence, minimum value of $v$ for which the particle will not hit the screen.
$v=\frac{q B_{0} d}{m}$
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