An electric field due to a positively charged | vedclass

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(B) The electron revolves around the wire due to the electrostatic force of attraction. Given that the electric field $E \propto r^{-1}$,we can write

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$1: 2$

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(B) The electron revolves around the wire due to the electrostatic force of attraction. Given that the electric field $E \propto r^{-1}$,we can write $E = k r^{-1}$,where $k$ is a constant.The electrostatic force on the electron is $F = e E = k e r^{-1}$.This force provides the necessary centripetal force for the circular motion of the electron:$\frac{m v^2}{r} = \frac{k e}{r}$Solving for the velocity $v$:$v^2 = \frac{k e}{m} \Rightarrow v = \sqrt{\frac{k e}{m}}$Since the velocity $v$ is independent of the radius $r$,the speed of the electrons in both orbits is the same,i.e.,$v_1 = v_2$.The time period $T$ of an orbit is given by $T = \frac{2 \pi r}{v}$.Therefore,the ratio of the time periods for radii $r_1 = 1 \mathring{A}$ and $r_2 = 2 \mathring{A}$ is:$\frac{T_1}{T_2} = \frac{2 \pi r_1 / v_1}{2 \pi r_2 / v_2} = \frac{r_1}{r_2} = \frac{1 \mathring{A}}{2 \mathring{A}} = \frac{1}{2}$.

An electric field due to a positively charged long straight wire at a distance $r$ from it is proportional to $r^{-1}$ in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii $1 \mathring{A}$ and $2 \mathring{A}$. The ratio of their respective time periods is

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