JEE Main 2025 Jan 28 Shift 1, Physics Q6: Motion of a Charged Particle in a Magnetic Field

JEE Main2025Jan 28, Shift 1Physics

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 [ $μ_{0}$ is vacuum permeability]

A
$a [1 - \frac{m v _{0}}{2 q μ _{0} I}]$
B
$\frac{a}{2}$
C
$a [1 - \frac{m v _{0}}{q μ _{0} I}]$
D
$a e^{- \frac{4 π m v _{0}}{q μ _{0} I}}$

Solution

Magnetic field at a distance $r$ from the long straight wire is

$B = \frac{μ _{0} I}{2 π r}$

The magnetic force on the particle is given by

$F = q (v \times B)$

Resolving the motion into $x$- and $y$-components,

$M \frac{d v _{x}}{d t} = q v_{y} B$

$M \frac{d v _{y}}{d t} = - q v_{x} B$

Since $v_x=\dfrac{dr}{dt}$,

$M \frac{d v _{y}}{d r} = - \frac{μ _{0} I q}{2 π r}$

Integrating from $r=a,\;v_y=v_0$ to $r=x,\;v_y=-v_0$,

$\int_{v_{0}}^{- v_{0}} M d v_{y} = - \frac{μ _{0} I q}{2 π} \int_{a}^{x} \frac{d r}{r}$

$- 2 M v_{0} = - \frac{μ _{0} I q}{2 π} ln (\frac{x}{a})$

$ln (\frac{a}{x}) = \frac{4 π M v _{0}}{μ _{0} I q}$

Hence,

$x = a exp (- \frac{4 π M v _{0}}{μ _{0} I q})$

Concept behind this question

Motion of a Charged Particle in a Magnetic FieldNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 1, Physics Q6: Motion of a Charged Particle in a Magnetic Field

Concept behind this question

More previous year questions on Motion of a Charged Particle in a Magnetic Field

Practice more Physics