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(B) The magnetic field produced by a moving charge $q$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} 	imes \vec{r})}{r^3}$.The magnetic f

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$\frac{2\pi}{3t}$

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(B) The magnetic field produced by a moving charge $q$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} 	imes \vec{r})}{r^3}$.The magnetic field at point $A$ is zero when the velocity vector $\vec{v}$ of the $\alpha$ particle is parallel or anti-parallel to the position vector $\vec{r}$ (the vector from the particle to point $A$).This happens when the particle is at the points of tangency on the circle as seen from point $A$.Let $	heta$ be the angle between the line connecting the center to point $A$ and the radius to the point of tangency.From the geometry,$\cos 	heta = \frac{R}{2R} = \frac{1}{2}$,so $	heta = \frac{\pi}{3}$.The particle moves from one tangent point to the other through the arc corresponding to an angle of $2	heta = \frac{2\pi}{3}$.The time taken to cover this angular displacement is $t = \frac{2	heta}{\omega} = \frac{2\pi/3}{\omega} = \frac{2\pi}{3\omega}$.Therefore,$\omega = \frac{2\pi}{3t}$.

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