A particle of mass m and charge q,accelerated | vedclass

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Question

(A) Let $v$ be the velocity of the particle. Its kinetic energy is given by:$\frac{1}{2}mv^2 = qV \implies v = \sqrt{\frac{2qV}{m}} \quad ... (1)$The

Vedclass · Accepted Answer

$\sin 	heta = Bd{\left( {\frac{q}{{2mV}}} ight)^{\frac{1}{2}}}$

Vedclass · Answer

(A) Let $v$ be the velocity of the particle. Its kinetic energy is given by:$\frac{1}{2}mv^2 = qV \implies v = \sqrt{\frac{2qV}{m}} \quad ... (1)$The particle follows a circular path of radius $r$ in the magnetic field,where:$\frac{mv^2}{r} = qvB \implies r = \frac{mv}{qB} \quad ... (2)$Substituting the value of $v$ from equation $(1)$ into equation $(2)$:$r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$From the geometry of the path shown in the figure,in the right-angled triangle formed,the angle of deviation $	heta$ satisfies:$\sin 	heta = \frac{d}{r}$Substituting the expression for $r$:$\sin 	heta = \frac{d}{\frac{1}{B} \sqrt{\frac{2mV}{q}}} = Bd \sqrt{\frac{q}{2mV}} = Bd \left( \frac{q}{2mV} ight)^{1/2}$Thus,the correct option is $(A)$.

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