Physics - Magnetism Question with Solution | TestHub

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