Consider a long thin conducting wire carrying a uniform current I. A particle having mass “M” and charge “q” is released at a distance “a" from the wire with a speed v_o along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [ $μ$ ₀ is vacuum permeability]

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$a [1 - \frac{{mv}_{o}}{2 q μ_{o} I}]$

$a e^{\frac{- 4 π m v_{o}}{q μ_{o} I}}$

$a [1 - \frac{{mv}_{o}}{q μ_{o} I}]$

$\frac{a}{2}$

answer is D.

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Detailed Solution

$\begin{aligned} A \to B \\ \vec{V} = - v_{x} \hat{i} + v_{y} \hat{j} \\ \vec{B} = \frac{μ_{0} I}{2 π r} (- \hat{k}) \\ \vec{F} = q (\vec{v} \times \vec{B}) = \frac{μ_{0} Iq}{2 π r} [- v_{x} \hat{j} - v_{y} \hat{i}] \\ a_{x} = - \frac{μ_{0} Iq}{2 π m} \cdot \frac{v_{y}}{r} \\ a_{y} = - \frac{μ_{0} I q}{2 π m} \cdot \frac{v_{x}}{r} \\ \frac{v_{x} {dv}_{x}}{dr} = - \frac{μ_{0} Iq}{2 π m} \frac{v_{y}}{r} \\ \frac{v_{x} {dv}_{x}}{v_{y}} = - \frac{μ_{0} Iq}{2 π m} \frac{dr}{r} \\ \int_{0}^{v_{0}} \frac{v_{x} d v_{x}}{\sqrt{v_{0}^{2} - v_{x}^{2}}} = - \frac{μ_{0} I q^{x_{1}}}{2 π m} \int_{a}^{d r} \frac{d r}{r} \\ Let, z^{2} = {v_{0}}^{2} - v_{x}^{2} \\ 2 zdz = - 2 v_{x} {dvx}_{x} \end{aligned}$