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A particle of charge Q and mass m moves in a circular path of radius R in a uniform magnetic field of magnitude B. The same particle now moves with the same speed in a circular path of same radius R in the space between the cylindrical electrodes of the cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $3R/2$ . Then the potential difference between the capacitor electrodes must be

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QBR (In3) /m

Zero

{QB}^{2} R^{2} (In3) /2m

{QB}^{2} R^{2} (In3) /m

answer is C.

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Detailed Solution

$F_{ER} = F_{CP}$

$\Rightarrow E_{q} = \frac{{mV}^{2}}{r}$

$R = \frac{mV}{Bq} V = \frac{BqR}{m}$

$\frac{λ Q}{2 π E_{0} R} = \frac{m}{R} {(\frac{BRQ}{m})}^{2}$

$\frac{λ}{2 π E_{0}} = \frac{B^{2} R^{2} Q}{m} \to (1)$

$dV = Edr \Rightarrow dV = \frac{λ}{2 π E_{0}} . \frac{1}{r} dr$

$V_{2} - V_{1} = \frac{λ}{2 π E_{0}} {(\ln r)}_{8 / 2}^{3R/2}$

$V_{2} - V_{1} = \frac{B^{2} R^{2} Q}{m} (\ln 3)$

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