A light charged particle is revolving in a ci | vedclass

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(D) The centripetal force required for circular motion is provided by the electrostatic force of attraction:$F_{	ext{centripetal}} = m r \omega^{2} =

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

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(D) The centripetal force required for circular motion is provided by the electrostatic force of attraction:$F_{	ext{centripetal}} = m r \omega^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$From this, we find the angular velocity $\omega$:$\omega^{2} = \frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} m r^{3}} \implies \omega \propto \frac{1}{r^{3/2}}$The moving charge $q_{1}$ constitutes a current $i$ given by:$i = \frac{q_{1}}{T} = \frac{q_{1} \omega}{2 \pi}$The magnetic field $B$ at the center of a circular current loop is:$B = \frac{\mu_{0} i}{2 r} = \frac{\mu_{0}}{2 r} \left( \frac{q_{1} \omega}{2 \pi} ight) = \frac{\mu_{0} q_{1} \omega}{4 \pi r}$Since $\omega \propto r^{-3/2}$, we have:$B \propto \frac{\omega}{r} \propto \frac{r^{-3/2}}{r} = r^{-5/2}$Thus, $B \propto \frac{1}{r^{5/2}}$.

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

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