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(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.The radius $R$ of this circular p

Vedclass · Accepted Answer

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

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(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.The radius $R$ of this circular path is given by $R = \frac{mv}{qB_{0}}$.For the particle not to hit the screen placed at a distance $d$ from its initial position,the radius of the circular path must be less than or equal to $d$.Therefore,$R \leq d$.Substituting the expression for $R$,we get $\frac{mv}{qB_{0}} \leq d$.Solving for $v$,we get $v \leq \frac{q B_{0} d}{m}$.However,the question asks for the minimum value of $v$ such that the particle does not hit the screen. If the particle's path is tangent to the screen,it will just graze it without hitting it. This occurs when $R = d$.Thus,$v = \frac{q B_{0} d}{m}$ is the required value.

$A$ 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 $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

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