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An electric current $I$ enters and leaves a uniform circular wire of radius $r$ through diametrically opposite points. A charged particle $q$ moves along the axis of circular wire passes through its centre with speed $v$. The magnetic force on the particle when it passes through the centre has a magnitude
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Force on a moving charged particle in uniform magnetic field,
$$
F=B q u \sin \theta
$$
Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^{\circ}$ or $180^{\circ}$.
When $\theta^{\circ}=0^{\circ}, F=B q v \sin 0^{\circ}$
[from Eq. (i)]
$$
\begin{aligned}
& F=0 \\
& \text { When } \theta=180^{\circ}, F=B q v \sin 180^{\circ} \\
& F=0 \\
&
\end{aligned}
$$
$$
F=B q u \sin \theta
$$
Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^{\circ}$ or $180^{\circ}$.
When $\theta^{\circ}=0^{\circ}, F=B q v \sin 0^{\circ}$
[from Eq. (i)]
$$
\begin{aligned}
& F=0 \\
& \text { When } \theta=180^{\circ}, F=B q v \sin 180^{\circ} \\
& F=0 \\
&
\end{aligned}
$$
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