JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2025 · 28 January · Morning Shift · Physics · Magnetic Effects of Current

Consider a long thin conducting wire carrying a uniform current I....

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

q

" is released at a distance "

a

" from the wire with a speed

v_0

along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance

x

from the wire. The value of

x

is [

\mu_0

is vacuum permeability]

Options

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

Correct answer

A. $a e^{-\frac{4 \pi \mathrm{mv}_{\mathrm{o}}}{\mathrm{q} \mu_{\mathrm{o}}$ \mathrm{I} $}}$

Step-by-step 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

\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}}}$

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