JEE Main 2025 — Magnetic Effects of Current Question with Solution

$\begin{array}{rl}& A\to B\\ & \vec{V}=-v_x\hat{i}+v_y\hat{j}\\ & \vec{B}=\frac{{\mu }_{0}I}{2\pi r}(-\hat{k})\\ & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})=\frac{{\mu }_0\mathrm{Iq}}{2\pi r}\left[-{\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{j}}-{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{i}}\right]\\ & {\mathrm{a}}_{\mathrm{x}}=-\frac{{\mu }_0\mathrm{Iq}}{2\pi \mathrm{m}}\cdot \frac{v_y}{\mathrm{r}}\\ & {\mathrm{a}}_{\mathrm{y}}=-\frac{{\mu }_0\mathrm{Iq}}{2\pi m}\cdot \frac{v_x}{\mathrm{r}}\end{array}$ $\begin{array}{rl}& \frac{{\mathrm{v}}_{\mathrm{x}}{\mathrm{dv}}_{\mathrm{x}}}{\mathrm{dr}}=-\frac{{\mu }_0\mathrm{Iq}}{2\pi \mathrm{m}}\frac{{\mathrm{v}}_{\mathrm{y}}}{\mathrm{r}}\\ & \frac{{\mathrm{v}}_{\mathrm{x}}{\mathrm{dv}}_{\mathrm{x}}}{{\mathrm{v}}_{\mathrm{y}}}=-\frac{{\mu }_0\mathrm{Iq}}{2\pi \mathrm{m}}\frac{\mathrm{dr}}{\mathrm{r}}\\ & {\int }_0^{{\mathrm{v}}_0}\frac{{\mathrm{v}}_{\mathrm{x}}{\mathrm{dv}}_{\mathrm{x}}}{\sqrt{{\mathrm{v}}_0^2-{\mathrm{v}}_{\mathrm{x}}^2}}=-\frac{{\mu }_0{\mathrm{Iq}}^{{\mathrm{x}}_1}}{2\pi \mathrm{m}}{\int }_{\mathrm{a}}^{\mathrm{dr}}\frac{\mathrm{r}}{2}\\ & \text{ Let, }{\mathrm{z}}^2={\mathrm{v}}_0^2-{\mathrm{v}}_{\mathrm{x}}^2\\ & 2\mathrm{zdz}=-2{\mathrm{v}}_{\mathrm{x}}{\mathrm{dv}}_{\mathrm{x}}\end{array}$ $\begin{aligned} & \mathrm{zdz}=-\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}} \\ & \frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_x}{\sqrt{\mathrm{v}_0^2-\mathrm{v}_{\mathrm{x}}^2}}=\frac{-\mathrm{zdz}}{\mathrm{z}}=-\mathrm{dz} \end{aligned}$ then integral becomes $\begin{aligned} & -\int_{\mathrm{v}_0}^0 \mathrm{dz}=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_1}{\mathrm{a}} \\ & \mathrm{v}_0=-\frac{\mu_0 \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_1}{\mathrm{a}} \end{aligned}$ ${\mathrm{x}}_1=\mathrm{a}{\mathrm{e}}^{-\frac{2\pi \mathrm{mv}}{{\mu }_0}}{\mathrm{H}}_0\mathrm{q}|\dots \dots$ For B $\to$ 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 $X=a{e}^{-\frac{4\pi {\mathrm{mv}}_0}{{\mu }_0\mathrm{Iq}}}$