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A charged particle moving in $x - y$ plane enters a region of uniform magnetic field of intensity $\overset{⇀}{B} = - B_{0} \hat{k}$ at $P (0, 0)$ with velocity ${\overset{⇀}{v}}_{1} = v \hat{i}$ and leaves the magnetic field at $Q (a, b)$ with velocity $\overset{⇀}{v_{2}} = v' \hat{i} + \sqrt{3} v' \hat{j}$ . Choose the correct statement(s):

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$\overset{⇀}{v_{1}} = \overset{⇀}{v_{2}}$

$\sqrt{3} a = b$

The particle can't be positron

Deviation suffered by the particle as it moves from $P$ to $Q$ equals $60 °$

answer is B.

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Detailed Solution

$\tan θ = \frac{v_{y}}{v_{x}} = \frac{\sqrt{3}}{1} = \sqrt{3}$

or $θ = 60 °$

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