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1. Electric Charges and Fields
medium
An electron revolves around an infinite cylindrical wire having uniform linear change density $2 \times 10^{-8}\,Cm ^{-1}$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is $.........\times 10^6\,ms ^{-1}$. Given mass of electron $=9 \times 10^{-31}\,kg$
A
$4$
B
$2$
C
$8$
D
$6$
(JEE MAIN-2023)
Solution
$e E=\frac{ mV ^2}{ r }$
$e \cdot \frac{2 K \lambda}{ r }=\frac{ mV ^2}{ r }$
$V =\sqrt{\frac{e \cdot 2 k \lambda}{ m }}$
$=\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}}$
$=8 \times 10^6\,m / s$
Standard 12
Physics
