# A particle of change 'q' and mass 'm' is moving with a velocit $-vi(v\neq 0)$ towards a large screen in the Y-Z plane at a distance 'd'. if there is a magnetic field $\overrightarrow{B}=Bo\widehat{k}$, the minimum value of v for which the particle will not hit the screen is: Option: 1 Option: 2 Option: 3 Option: 4

As velocity and magnetic field is perpendicular to each other So particle will perform uniform circular motion in x-y plane

And its radius is given as

$r=\frac{m v}{q B_0}$

So if  the particle will not hit the screen then

$d\frac{dq B_0}{m}\\ V_{min}=\frac{qd B_0}{m}$

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