Question
A particle of charge $$-q$$ and mass $$m$$ moves in a circle of radius around an infinitely long line charge of linear charge density $$ + \lambda .$$ Then time period will be
A particle of charge $$-q$$ and mass $$m$$ moves in a circle of radius around an infinitely long line charge of linear charge density $$ + \lambda .$$ Then time period will be
A.
$$T = 2\pi r\sqrt {\frac{m}{{2k\lambda q}}} $$
B.
$${T^2} = \frac{{4{\pi ^2}m}}{{2k\lambda q}}{r^3}$$
C.
$$T = \frac{1}{{2\pi r}}\sqrt {\frac{{2k\lambda q}}{m}} $$
D.
$$T = \frac{1}{{2\pi r}}\sqrt {\frac{m}{{2k\lambda q}}} $$
Answer :
$$T = 2\pi r\sqrt {\frac{m}{{2k\lambda q}}} $$
Solution :
We have centripetal force equation
$$\eqalign{ & q\left( {\frac{{2k\lambda }}{r}} \right) = \frac{{m{v^2}}}{r} \cr & {\text{so}}\,\,v = \sqrt {\frac{{2kq\lambda }}{m}} \cr & {\text{Now,}}\,\,T = \frac{{2\pi r}}{v} = 2\pi r\sqrt {\frac{m}{{2kq\lambda }}} \cr & {\text{where}}\,\,k = \frac{1}{{4\pi {\varepsilon _0}}} \cr} $$
We have centripetal force equation
$$\eqalign{ & q\left( {\frac{{2k\lambda }}{r}} \right) = \frac{{m{v^2}}}{r} \cr & {\text{so}}\,\,v = \sqrt {\frac{{2kq\lambda }}{m}} \cr & {\text{Now,}}\,\,T = \frac{{2\pi r}}{v} = 2\pi r\sqrt {\frac{m}{{2kq\lambda }}} \cr & {\text{where}}\,\,k = \frac{1}{{4\pi {\varepsilon _0}}} \cr} $$
