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(D) The magnetic force on a moving charge is given by $\vec F = q(\vec v 	imes \vec B)$.If $\vec v 	imes \vec B = \vec 0$,the magnetic force is zero

Vedclass · Accepted Answer

Time period $T$ does not depend on the angle between $\vec v$ and $\vec B$.

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(D) The magnetic force on a moving charge is given by $\vec F = q(\vec v 	imes \vec B)$.If $\vec v 	imes \vec B = \vec 0$,the magnetic force is zero,and the particle continues to move in a straight line.If $\vec v$ is perpendicular to $\vec B$,the path is circular with time period $T = \frac{2\pi m}{qB}$.If $\vec v$ makes an angle $	heta$ with $\vec B$,the velocity component perpendicular to the field is $v_{\perp} = v \sin 	heta$.The radius of the circular path is $r = \frac{mv \sin 	heta}{qB}$.The time period is $T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi (mv \sin 	heta / qB)}{v \sin 	heta} = \frac{2\pi m}{qB}$.Thus,the time period $T$ is independent of the angle $	heta$ between $\vec v$ and $\vec B$.

$A$ particle of mass $m$ and charge $q$ moving with velocity $\vec v$ enters a region of uniform magnetic field $\vec B.$ Then:

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