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A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\overrightarrow{ u }_1=4 \hat{ i } ms ^{-1}$, enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x=0$ to $x$ $=L$ for all values of $y$. After passing through this region, the particle emerges on the other side after $10$ milliseconds with a velocity $\overline{ u }_2=2(\sqrt{3} \hat{ i }+\hat{ j }) ms ^{-1}$. The correct statement$(s)$ is (are) :
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction
$(C)$ The magnitude of the magnetic field $\frac{50 \pi M }{3 Q }$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100 \pi M}{3 Q}$ units.
$(B,D)$
$(B,C)$
$(A,C)$
$(A,D)$
Solution
Component of final velocity of particle is in positive y direction.
Centre of circle is present on positive y axis. so magnetic field is present in negative $z$-direction Angle of deviation is $30^{\circ}$ because
$\tan \theta=\frac{v_y}{v_x}=\frac{1}{\sqrt{3}} $
$\theta=\frac{\pi}{6} $
$\omega t=\theta $
$\theta=\frac{Q B}{M} t $
$B=\frac{M \theta}{Q t} $
$B=\left(\frac{50 M \pi}{3 Q}\right)$
