For $$A \to B$$

$$\overrightarrow V = - {v_x}\widehat i + {v_y}\widehat j$$

$$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$$

$\because$ $$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$$

So, $$\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ {\left[ {\left( { - {v_x}} \right)\widehat i + {v_y}\widehat j} \right] \times \left( { - \widehat k} \right)} \right]$$

$$ = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) - {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$$

$$\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ { - {v_x}\widehat j - {v_y}\widehat i} \right]$$

$$ \Rightarrow {F_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$$

$$ \Rightarrow m{a_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$$

$$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$$

$$ \Rightarrow {{{v_x}d{v_x}} \over {{v_y}}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{dr} \over r}$$

Since, $$v_x^2 + v_y^2 = v_0^2$$

$$ \Rightarrow {v_y} = \sqrt {v_0^2 - v_x^2} $$

$$ \Rightarrow \int_0^{{v_0}} {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{ - {\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $$

Let $$v_0^2 - v_x^2 = {z^2}$$

$$ \Rightarrow - 2{v_x}d{v_x} = 2zdz$$

when $${v_x} = 0,\,z = {v_0}$$

$${v_x} = {v_0},\,z = 0$$

$$ \Rightarrow \int_{{v_0}}^0 {{{zdz} \over z} = - {{{\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $$

$$ \Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{{x'} \over a}} \right)$$

$$ \Rightarrow \ln \left( {{{x'} \over a}} \right) = - {{2\pi m{v_0}} \over {{\mu _0}Iq}}$$

$$ \Rightarrow x' = ac{{ - 2\pi m{v_0}} \over {{\mu _0}Iq}}$$ .... (1)

Now, for $$B \to C$$

$$\overrightarrow v = - {v_x}\widehat i - {v_y}\widehat j$$

$$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$$

$$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) + {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$$

$$ \Rightarrow \overrightarrow F = {{{\mu _0}Iq} \over {2\pi r}}\left[ { - {v_x}\widehat j + {v_y}\widehat i} \right]$$

$$ \Rightarrow {F_x} = {{{\mu _0}Iq} \over {2\pi r}}{v_y}$$

$$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{{\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$$

$$\int_{{v_0}}^0 {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{{\mu _0}Iq} \over {2\pi m}}\int_{x'}^x {{{dr} \over r}} } $$

Let $$v_0^2 - v_x^2 = {z^2}$$

$$ - 2{v_x}d{v_x} = 2zdz$$

$${v_x} = {v_0} \Rightarrow z = 0$$

$${v_x} = 0 \Rightarrow z = {v_0}$$

$$ \Rightarrow - \int_0^{{v_0}} {{{zdz} \over z} = {{{\mu _0}Iq} \over {2\pi m}}\left[ {\ln r} \right]_{x'}^x} $$

$$ \Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{x \over {x'}}} \right)$$

$$ \Rightarrow x = x'{e^{ - {{2\pi m{v_0}} \over {{\mu _0}Iq}}}}$$ .... (2)

From (1) & (2), $$x = a{e^{ - {{4\pi m{v_0}} \over {{\mu _0}Iq}}}}$$