Consider a long thin conducting wire carrying a uniform current I. A particle having mass " M " and charge "…

1. $a e^{-\frac{4 \pi \mathrm{mv}_{\mathrm{o}}}{\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}}}$
2. $a\left[1-\frac{\mathrm{mv}_{\mathrm{o}}}{2 q \mu_{\mathrm{o}} \mathrm{I}}\right]$
3. $a\left[1-\frac{\mathrm{mv}}{\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}}\right]$
4. $\frac{a}{2}$

Solution

$\begin{aligned} & A \rightarrow B \\ & \vec{V}=-v_x \hat{i}+v_y \hat{j} \\ & \vec{B}=\frac{\mu_0 I}{2 \pi r}(-\hat{k}) \\ & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_0 \mathrm{Iq}}{2 \pi r}\left[-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{j}}-\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}\right] \\ & \mathrm{a}_{\mathrm{x}}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{v_y}{\mathrm{r}} \\ & \mathrm{a}_{\mathrm{y}}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi m} \cdot \frac{v_x}{\mathrm{r}}\end{aligned}$
$\begin{aligned} & \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}} \\ & \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{y}}}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{dr}}{\mathrm{r}} \\ & \int_0^{\mathrm{v}_0} \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_0^2-\mathrm{v}_{\mathrm{x}}^2}}=-\frac{\mu_0 \mathrm{Iq}^{\mathrm{x}_1}}{2 \pi \mathrm{~m}} \int_{\mathrm{a}}^{\mathrm{dr}} \frac{\mathrm{r}}{2} \\ & \text { Let, } \mathrm{z}^2=\mathrm{v}_0^2-\mathrm{v}_{\mathrm{x}}^2 \\ & 2 \mathrm{zdz}=-2 \mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}\end{aligned}$
$\begin{aligned}
& \mathrm{zdz}=-\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}} \\ & \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_x}{\sqrt{\mathrm{v}_0^2-\mathrm{v}_{\mathrm{x}}^2}}=\frac{-\mathrm{zdz}}{\mathrm{z}}=-\mathrm{dz}
\end{aligned}$
then integral becomes
$\begin{aligned}
& -\int_{\mathrm{v}_0}^0 \mathrm{dz}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_1}{\mathrm{a}} \\ & \mathrm{v}_0=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_1}{\mathrm{a}}
\end{aligned}$
$\left.\mathrm{x}_1=\mathrm{a} \mathrm{e}^{-\frac{2 \pi \mathrm{mv}}{\mu_0}} \mathrm{H}_0 \mathrm{q} \right\rvert\, \ldots \ldots$
For B $\rightarrow$ C
$\begin{aligned}
& \overrightarrow{\mathrm{v}}=-v_x \hat{\mathrm{i}}-v_y \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{~B}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_0 \mathrm{Iq}}{2 \pi r}\left(-v_x \hat{j}+v_y \hat{\mathrm{i}}\right)
\end{aligned}$
$\begin{aligned}
& \mathrm{a}_{\mathrm{x}}=+\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}} \quad \mathrm{a}_{\mathrm{y}}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{r}} \\ & \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}} \\ & \int_{v_0}^0 \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_0^2-\mathrm{v}_{\mathrm{x}}^2}}=\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \int_{\mathrm{x_1}}^{\mathrm{x}}\frac{dr}{r}
\end{aligned}$
$\begin{aligned}
& \frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}}{\mathrm{x}_1}=-\int_0^{\mathrm{v}_0} \mathrm{~d} \mathrm{z}=-\mathrm{v}_0 \\ & \mathrm{x}=\mathrm{x}_1 \mathrm{e}^{-\frac{2 \pi \mathrm{~m}_0}{\mu_0 \mathrm{Iq}}} \ldots \ldots \text { (2) }
\end{aligned}$
From equation 1 and 2
$X=a e^{-\frac{4 \pi \mathrm{mv}_0}{\mu_0 \mathrm{Iq}}}$

Asked in: JEE Main 2025 (28 Jan Shift 1)

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