Two particles A and B having equal charges +6 | vedclass

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Question

(A) Let $v$ be the velocity acquired by the charged particle when accelerated through the potential difference $V$.$\frac{1}{2}mv^2 = qV \implies v =

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(A) Let $v$ be the velocity acquired by the charged particle when accelerated through the potential difference $V$.$\frac{1}{2}mv^2 = qV \implies v = \sqrt{\frac{2qV}{m}}$When a charged particle enters a uniform magnetic field $B$ perpendicular to its velocity,it describes a circular path of radius $R$ given by:$R = \frac{mv}{qB}$Substituting the value of $v$:$R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$Since $q, V,$ and $B$ are the same for both particles,we have $R \propto \sqrt{m}$.Therefore,$\frac{R_A}{R_B} = \sqrt{\frac{m_A}{m_B}}$.Squaring both sides:$\frac{m_A}{m_B} = \left(\frac{R_A}{R_B}\right)^2 = \left(\frac{2\,cm}{3\,cm}\right)^2 = \frac{4}{9}$.

Two particles $A$ and $B$ having equal charges $+6\,C$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $2\,cm$ and $3\,cm$ respectively. The ratio of mass of $A$ to that of $B$ is

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