A positive point charge ‘q’ of mass ‘m’ kept at a distance $x_{0}$ (in the same plane) from a fixed very long straight wire carrying current ‘I’ is projected normally away from it with speed ‘v’ (in the same plane). Which of the following statements is correct for subsequent motion (ignore gravity):

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The velocity of the particle remains constant

The angular momentum of the particle along the wire does not change as the charge moves

The maximum separation of the particle from the wire is given by $x = x_{0} e^{2 p m v / m_{0} q i}$

The radius of curvature of the path of particle remains constant

answer is B, C.

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Detailed Solution

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$\vec{F} = q \vec{v} \times \vec{β} = q (v_{x} \overset{\land}{i} + v_{y} \overset{\land}{j}) \times \frac{μ_{0} I}{2 π x} (- \overset{\land}{k}) \Rightarrow a_{x} = \frac{- μ_{0} q I v_{y}}{2 π x m}, a_{y} = \frac{μ_{0} I q v_{x}}{2 π x m} A l s o, v_{x}^{2} + v_{y}^{2} = v^{2}$

Differentiate: $2 v_{x} d v_{x} + 2 v_{y} d v_{y} = 0$

$\Rightarrow v_{x} d v_{x} = - v_{y} d v_{y} \Rightarrow \frac{2 π m}{μ_{0} q I} = d v_{y} = \frac{d x}{x} & a s x_{\max}, v_{x} = 0$

$\Rightarrow v_{y} = v \Rightarrow \frac{2 π m}{μ_{0} q I} \int_{0}^{v} d v_{y} = {[l n x]}_{x_{0}}^{x_{m}} \Rightarrow x_{m} = x_{0} e^{\frac{2 π m v}{μ_{0} g I}} A n s . B C$