Two particles, A and B, having equal charges, | vedclass

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(B) The radius $R$ of the circular path followed by a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, wh

Vedclass · Accepted Answer

$(R_{1} / R_{2})^{2}$

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(B) The radius $R$ of the circular path followed by a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, where $K$ is the kinetic energy.Since the particle is accelerated through a potential difference $V$, its kinetic energy is $K = qV$.Substituting this into the radius formula, we get $R = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Given that $q$, $V$, and $B$ are the same for both particles, we have $R \propto \sqrt{m}$.Therefore, $\frac{R_{1}}{R_{2}} = \sqrt{\frac{m_{A}}{m_{B}}}$.Squaring both sides, we get $\frac{m_{A}}{m_{B}} = \left(\frac{R_{1}}{R_{2}}\right)^{2}$.

Two particles, $A$ and $B$, having equal charges, after being accelerated through the same potential difference, enter into a region of uniform magnetic field and the particles describe circular paths of radii $R_{1}$ and $R_{2}$ respectively. The ratio of the masses of $A$ and $B$ is

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