Two charged particles, having same kinetic energy are allowed to pass through uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6 : 7$ and their respective masses ratio is $9 : 4$ Then, the ratio of their charges will be

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7:4

8:7

8:5

5:3

answer is B.

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Detailed Solution

$G i v e n K_{A} = K_{B}$

$\begin{array}{l} \frac{1}{2} m_{A} V_{A}^{2} = \frac{1}{2} m_{B} V_{B}^{2} \\ & \frac{m_{A}}{m_{B}} = \frac{9}{4} a l s o \frac{r_{A}}{r_{B}} = \frac{6}{7} \\ n o w \frac{1}{2} m_{A} V_{A}^{2} = \frac{1}{2} m_{B} V_{B}^{2} \end{array}$

$\begin{array}{l} \Rightarrow {(\frac{V_{A}}{V_{B}})}^{2} = \frac{m_{B}}{m_{A}} = \frac{4}{9} \\ o r \frac{V_{A}}{V_{B}} = \frac{2}{3} \end{array}$

$\begin{array}{l} \sin c e r = \frac{m V}{B q} \\ \frac{r_{A}}{r_{B}} = \frac{m_{A}}{m_{B}} \frac{V_{A}}{V_{B}} \times \frac{B}{B} \frac{q_{B}}{q_{A}} \end{array}$

On substituting values, we get

$\begin{array}{l} \frac{6}{7} = \frac{9}{4} \times \frac{2}{3} \times \frac{q_{B}}{q_{A}} \\ \Rightarrow \frac{q_{A}}{q_{B}} = \frac{7}{4} \end{array}$