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(B) The particle enters a uniform magnetic field $B$ perpendicular to its velocity and will follow a circular path.The radius $r$ of the circular path

Vedclass · Accepted Answer

$B > \frac{\sqrt{2mE}}{qd}$

Vedclass · Answer

(B) The particle enters a uniform magnetic field $B$ perpendicular to its velocity and will follow a circular path.The radius $r$ of the circular path is given by $r = \frac{mv}{qB}$,where $v$ is the velocity of the particle.Given the kinetic energy $E = \frac{1}{2}mv^2$,we can write $v = \sqrt{\frac{2E}{m}}$.Substituting $v$ into the radius formula,we get $r = \frac{m}{qB} \sqrt{\frac{2E}{m}} = \frac{\sqrt{2mE}}{qB}$.For the particle not to collide with the upper plate,the diameter of the circular path must be less than or equal to the distance $d$ between the plates. However,since the particle enters from the bottom plate and moves in a semicircle,the radius $r$ must be less than or equal to $d$ $(r \leq d)$.Therefore,$\frac{\sqrt{2mE}}{qB} \leq d$.Rearranging for $B$,we get $B \geq \frac{\sqrt{2mE}}{qd}$.Thus,the condition for the particle not to collide with the upper plate is $B > \frac{\sqrt{2mE}}{qd}$.

As shown in the figure,the uniform magnetic field between the two identical plates is $B$. There is a hole in the bottom plate. If a particle of charge $q$,mass $m$,and energy $E$ enters this magnetic field through this hole,then the particle will not collide with the upper plate provided:

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