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Question

(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity, it follows a circular path.
From the geometry of the path

Vedclass · Accepted Answer

$T \left( \frac{\pi - 2	heta}{2\pi} ight)$

Vedclass · Answer

(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity, it follows a circular path.
From the geometry of the path shown in the figure, the angle $\alpha$ subtended by the arc of the path inside the magnetic field at the center $C$ is related to the angle $	heta$ by the relation $\alpha + 	heta + 	heta = \pi$, which gives $\alpha = \pi - 2	heta$.
The time period of a full circular orbit is $T = \frac{2\pi m}{QB}$.
The time $t$ spent by the particle in the magnetic field is proportional to the angle $\alpha$ subtended at the center:
$t = \frac{\alpha}{2\pi} T$
Substituting $\alpha = \pi - 2	heta$, we get:
$t = \frac{\pi - 2	heta}{2\pi} T = T \left( \frac{\pi - 2	heta}{2\pi} ight)$.

$A$ particle with charge $-Q$ and mass $m$ enters a magnetic field of magnitude $B$, which exists only to the of the boundary $YZ$. The direction of the motion of the particle is perpendicular to the direction of $B$. Let $T = 2\pi \frac{m}{QB}$. The time spent by the particle in the field will be:

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