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 Y-Z plane 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 :

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$\frac{q d B_{0}}{2 m}$

$\frac{2 q d B_{0}}{m}$

$\frac{q d B_{0}}{3 m}$

$\frac{q d B_{0}}{m}$

answer is D.

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Detailed Solution

$T o a v o i d t h e h i t, r a d i u s o f c i r c u l a r m o t i o n s h o u l d b e j u s t g r e a t e r t h a n d i s \tan c e d . \Rightarrow r \geq d \Rightarrow \frac{m v}{q B_{0}} \geq d \Rightarrow ν_{m i n} = \frac{q B_{0} d}{m}$

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