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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 at speed $v$.
The magnetic force on the particle when it passes through the centre has a magnitude.
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The magnetic force on the particle when it passes through the centre has a magnitude.
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Verified Answer
The correct answer is:
0
0
Force on a moving charged particle in uniform magnetic field
$$
F=B q v \sin \theta
$$
Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^{\circ}$ or $180^{\circ}$
When $\theta=0^{\circ} ; F=B q v \sin 0^{\circ}$
$$
F=0
$$
When $\theta=180^{\circ}, F=B q u \sin 180^{\circ}$
$$
F=0
$$
$$
F=B q v \sin \theta
$$
Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta=0^{\circ}$ or $180^{\circ}$
When $\theta=0^{\circ} ; F=B q v \sin 0^{\circ}$
$$
F=0
$$
When $\theta=180^{\circ}, F=B q u \sin 180^{\circ}$
$$
F=0
$$
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