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(C) Let $R$ be the radius of the circular path of the charged particle in the magnetic field $B_0$. The particle enters perpendicular to side $AB$ and

Vedclass · Accepted Answer

$(2+\sqrt{3}) B_0$

Vedclass · Answer

(C) Let $R$ be the radius of the circular path of the charged particle in the magnetic field $B_0$. The particle enters perpendicular to side $AB$ and exits through side $BC$ with a deflection angle $	heta = 30^{\circ}$.From the geometry of the path,the distance $x$ from the side $BC$ to the point of entry is given by $R - x = R \cos 	heta$.Thus,$x = R(1 - \cos 30^{\circ}) = R(1 - \frac{\sqrt{3}}{2})$.When the magnetic field is changed to $B'$,the radius of the path becomes $R' = \frac{mv}{qB'}$. The deflection angle becomes $	heta' = 60^{\circ}$.The exit point $x$ remains the same as the geometry of the region is fixed. Thus,$R' - x = R' \cos 60^{\circ}$.$x = R'(1 - \cos 60^{\circ}) = R'(1 - \frac{1}{2}) = \frac{R'}{2}$.Equating the two expressions for $x$:$R(1 - \frac{\sqrt{3}}{2}) = \frac{R'}{2} \Rightarrow R' = 2R(1 - \frac{\sqrt{3}}{2}) = R(2 - \sqrt{3})$.Since $R = \frac{mv}{qB_0}$ and $R' = \frac{mv}{qB'}$,we have $\frac{1}{B'} = \frac{2 - \sqrt{3}}{B_0}$.$B' = \frac{B_0}{2 - \sqrt{3}} = B_0(2 + \sqrt{3})$.

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