Question

A light charged particle is resolving in a circle of radius 'r' in electrostatic attraction of a static heavy particle with opposite charge. How does the magnetic field 'B' at the centre of the circle due to the moving charge depend on 'r'?

• A. $B\propto \frac{1}{r}$
• B. $B\propto \frac{1}{r^2}$
• C. $B\propto \frac{1}{r^{\frac{3}{2}}}$
• D. $B\propto \frac{1}{r^{\frac{5}{2}}}$

Answer 1

Answer: (D)

$B\propto \frac{1}{r^{\frac{5}{2}}}$

Electrostatic force of attraction
$f=\frac{KqQ}{r^2}$
$\frac{m{v}^2}{r}=\frac{KqQ}{r^2}$
$v\propto \frac{1}{\sqrt{r}}$
T.P. $=\frac{2\pi r}{v}$
T.P. $\propto \frac{r}{v}$
T.P. $\propto r^{3/2}$
$I\propto \frac{Q}{T.P.}$
$\therefore I\propto r^{-3/2}$
$B=\frac{m{u}_{0}I}{2r}$
$B\propto I$
$B\propto \frac{I}{r}$
$B\propto \frac{I}{r}$ $B\propto \frac{r^{-3/2}}{2}$
$B\propto r^{-5/2}$