- Home
- Standard 12
- Physics
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 $1 A$ and $2 A$. The ratio of their respective time periods is
$1: 1$
$1: 2$
$2: 1$
$4: 1$
Solution
(b)
Electron revolve around wire due to its electrostatic force of attraction. As field $E \propto r^{-1} \Rightarrow E=k r^{-1}$
So, force on electron is
$F=e E=k e r^{-1}$
This force is necessary centripetal force.
So,
$\frac{m v^2}{r} =\frac{k e}{r}$
$\Rightarrow v =\sqrt{\frac{k e}{m}}$
As velocity of electron is independent of radius of paths,
$\Rightarrow v_1=v_2$
Now, ratio of time periods of rotation are
$\frac{T_1}{T_2}=\frac{\left(\frac{2 \pi r_1}{v_1}\right)}{\left(\frac{2 \pi r_2}{v_2}\right)}=\frac{r_1}{r_2}-\frac{1 \mathring A}{2 \mathring A}=\frac{1}{2}$
