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A rectangular region $A B C D$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. A narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $B C$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$, the magnetic field has to be
$\frac{3}{2} B_0$
$2 B _0$
$(2+\sqrt{3}) B _0$
$(3+\sqrt{3}) B_0$
Solution
(C)
$\cos \theta=\frac{(R-x)}{R} \Rightarrow \cos 30^{\circ}=1-\frac{x}{R}$
$\Rightarrow x=R\left(1-\frac{\sqrt{3}}{2}\right)$
Now $\cos \theta^{\prime}=\left(\frac{R^{\prime}-x}{R^{\prime}}\right)$
$\cos \theta^{\prime}=1-\frac{x}{R^{\prime}}$
$\cos 60^{\circ}=1-\frac{x}{R^{\prime}}$
$\frac{1}{2}=1-\frac{ R \left(1-\frac{\sqrt{3}}{2}\right)}{ R ^{\prime}}$
$R ^{\prime}=2 R \left(1-\frac{\sqrt{3}}{2}\right)$
$\frac{ mv }{ qB }=\frac{2 mv }{ qB }\left(1-\frac{\sqrt{3}}{2}\right)$
$B ^{\prime}=\frac{ B }{(2-\sqrt{3})}$
$B ^{\prime}=(2+\sqrt{3}) B _0$
