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(D) The magnetic force acting on a charged particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the force is al

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All of these

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(D) The magnetic force acting on a charged particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since the force is always perpendicular to the velocity $(\vec{F} \perp \vec{v})$,the magnetic field does no work on the particle. Therefore,the kinetic energy and the speed of the particle remain constant,so $v_1 = v_2$.Due to the symmetry of the circular path in a uniform magnetic field,the angle of entry $\alpha$ is equal to the angle of exit $\beta$,so $\alpha = \beta$.The particle follows a circular arc. The angle subtended by the arc at the center is $2(\pi - \alpha)$. The time spent in the magnetic field is $t = \frac{	heta}{\omega}$,where $\omega = \frac{qB}{m}$. Thus,$t = \frac{2(\pi - \alpha)m}{qB}$.Since all statements are correct,the correct option is $(D)$.

$A$ particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then

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