QJEEM25Ph4643: JEE Main 2025 Physics - Class 12 JEEM25Ph4643 Solution: Consider a long thin conducting wire carrying a uniform current …

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