Question

A small charged ball having mass m and charge q is suspended from a rigid support by means of an inextensible thread of length l. It is made to rotate on a horizontal circular path in a uniform, time independent magnetic field of induction B which is directed upward. The time period of revolution of the ball is $T_0$. If the radius of circular path on which the ball moves is $\therefore r={[l^2-\frac{(T_0/2\pi {)}^2}{\left\{\right(2\pi /g{T}_0)+(qB/mg\left)\right\}}]}^{1/x}$. Find x.

Assume string remain in stretched position.

Answer 1

The situation is shown in figure:
$Tcos\theta =mg$ (i)
$Tsin\theta -qvB=\left(\frac{m{v}^2}{r}\right)$ (ii)
From Eq. (ii),
$\left(\frac{mg}{cos\theta }\right)sin\theta -qBr\omega =mr{\omega }^2$
$[\because v=r\omega$ and $T=\frac{mg}{cos\theta }$ from equation (i)]
From figure, $sin\theta =\frac{r}{l}$ and $cos\theta =\frac{\sqrt{(l^2-r^2)}}{l}$
$\therefore \frac{mgr}{\sqrt{(l^2-r^2)}}-qBr\omega =mr{\omega }^2$
or $\frac{1}{\sqrt{(l^2-r^2)}}=\frac{{\omega }^2}{g}+\frac{qB\omega }{mg}$
or $(l^2-r^2)=\frac{(1/\omega {)}^2}{\left[\right(\omega /g)+(qB/mg{)}^2]}$
or $(l^2-r^2)=\frac{(T_0/2\pi {)}^2}{\left[\right(2\pi /g{T}_0)+(qB/mg){]}^2}$
$\therefore r={[l^2-\frac{(T_0/2\pi {)}^2}{\left\{\right(2\pi /g{T}_0)+(qB/mg\left)\right\}}]}^{1/2}$