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  4. A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only to the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of B. Then, the time spent by the particle in the field will be

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Q. A particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Then, the time spent by the particle in the field will be

NTA AbhyasNTA Abhyas 2020Moving Charges and Magnetism

Solution:

Solution
$\overset{ \rightarrow }{F}=Q\left(\overset{ \rightarrow }{V} \times \overset{ \rightarrow }{B}\right)$
$F=Q\left(V B\right) \, $
Since this force is constant and is perpendicular to particle's direction of motion. Particle will move in a circle.
$\therefore \, QVB=\frac{m V^{2}}{R}$
$R=\frac{m V}{Q B}$
$\therefore \, \, $ Time required to travel $\pi R$ distance
$t=\frac{\pi R}{V}$
$\Rightarrow \, t=\frac{\pi \times \frac{m V}{Q B}}{v}=\frac{\pi m}{Q B}$
$t=\frac{\pi m}{Q B}$