A particle of charge q,mass m enters a region | vedclass

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(A) As the initial velocity of the particle is perpendicular to the magnetic field,the particle moves along the arc of a circle.If $r$ is the radius o

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$\frac{mv_0}{2qB}$

Vedclass · Answer

(A) As the initial velocity of the particle is perpendicular to the magnetic field,the particle moves along the arc of a circle.If $r$ is the radius of the circular path,then the magnetic force provides the necessary centripetal force:$\frac{mv_0^2}{r} = qv_0B \Rightarrow r = \frac{mv_0}{qB}$From the geometry of the path,the particle enters at $x=0$ and exits at $x=d$. The angle of deviation is $30^{\circ}$.In the right-angled triangle formed by the radius $r$,the horizontal distance $d$,and the vertical component,we have:$\sin(30^{\circ}) = \frac{d}{r}$Substituting the value of $r$:$d = r \sin(30^{\circ}) = \left( \frac{mv_0}{qB} ight) 	imes \frac{1}{2} = \frac{mv_0}{2qB}$Thus,the correct option is $A$.

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

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