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(D) The magnetic field $\vec{B}$ due to the wire at a distance $r$ is $\vec{B} = \frac{\mu_0 I}{2 \pi r} (-\hat{k})$.As the particle moves,the magneti

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$ae^{-\frac{4 \pi m v_0}{q \mu_0 I}}$

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(D) The magnetic field $\vec{B}$ due to the wire at a distance $r$ is $\vec{B} = \frac{\mu_0 I}{2 \pi r} (-\hat{k})$.As the particle moves,the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ acts on it. Since the magnetic force is always perpendicular to the velocity,the speed $v_0$ remains constant.Let the velocity be $\vec{v} = -v_x \hat{i} + v_y \hat{j}$. The force is $\vec{F} = q(-v_x \hat{i} + v_y \hat{j}) 	imes \frac{\mu_0 I}{2 \pi r} (-\hat{k}) = \frac{\mu_0 I q}{2 \pi r} (-v_x \hat{j} - v_y \hat{i})$.The acceleration components are $a_x = -\frac{\mu_0 I q}{2 \pi m r} v_y$ and $a_y = -\frac{\mu_0 I q}{2 \pi m r} v_x$.Using $\frac{v_x dv_x}{dr} = a_x = -\frac{\mu_0 I q}{2 \pi m r} v_y$ and $v_y = \sqrt{v_0^2 - v_x^2}$,we get $\int_{0}^{v_0} \frac{v_x dv_x}{\sqrt{v_0^2 - v_x^2}} = -\frac{\mu_0 I q}{2 \pi m} \int_{a}^{x_1} \frac{dr}{r}$.Solving this gives $v_0 = \frac{\mu_0 I q}{2 \pi m} \ln(\frac{a}{x_1})$,so $x_1 = a e^{-\frac{2 \pi m v_0}{\mu_0 I q}}$.Due to symmetry,the particle turns back at $x = x_1 e^{-\frac{2 \pi m v_0}{\mu_0 I q}} = a e^{-\frac{4 \pi m v_0}{\mu_0 I q}}$.

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