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(D) The magnetic field is $\vec B = B \hat k$. The radius of the circular path of the particle is $r = \frac{mv}{qB}$.Given $d = \frac{mv}{2qB}$,we ha

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$\frac{qvB}{m} \left( \frac{1}{2} \hat j - \frac{\sqrt{3}}{2} \hat i \right)$

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(D) The magnetic field is $\vec B = B \hat k$. The radius of the circular path of the particle is $r = \frac{mv}{qB}$.Given $d = \frac{mv}{2qB}$,we have $r = 2d$.When the particle emerges at $y = d$,the angle $	heta$ that the velocity vector makes with the $x$-axis is given by $\sin 	heta = \frac{d}{r} = \frac{d}{2d} = \frac{1}{2}$,which implies $	heta = 30^\circ$ or $\frac{\pi}{6}$.The force on the particle is $\vec F = q(\vec v 	imes \vec B)$. The acceleration is $\vec a = \frac{\vec F}{m} = \frac{q}{m}(\vec v 	imes \vec B)$.At the point of emergence,the velocity vector is $\vec v = v(\cos 	heta \hat i + \sin 	heta \hat j)$.Thus,$\vec a = \frac{q}{m} [v(\cos 	heta \hat i + \sin 	heta \hat j) 	imes B \hat k] = \frac{qvB}{m} (\cos 	heta (-\hat j) + \sin 	heta (\hat i)) = \frac{qvB}{m} (\sin 	heta \hat i - \cos 	heta \hat j)$.Substituting $	heta = 30^\circ$,$\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we get $\vec a = \frac{qvB}{m} (\frac{1}{2} \hat i - \frac{\sqrt{3}}{2} \hat j)$.Comparing with the options,the magnitude and direction correspond to option $D$ (considering the sign of charge or direction of deflection).

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat k$. $A$ particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{mv}{2qB}$,the acceleration of the charged particle at the point of its emergence at the other side is

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