Question

A long straight wire carries a current $I$. A particle at point $P$ having a positive charge $q$ and mass $m$, kept at a distance $x_o$ from the wire, is projected towards it with a speed $v$ as shown in the diagram. Find the minimum separation between the wire and the particle.

$($Magnetic field due to wire is $\overrightarrow{B})$

• A. $x=x_o{e}^{-\left(\frac{mv}{{\mu }_{o}I}\right)}$
• B. $x=x_o{e}^{-\left(\frac{mv}{{\mu }_{o}qI}\right)}$
• C. $x=x_o{e}^{-\left(\frac{2\pi }{{\mu }_{o}I}\right)}$
• D. $x=x_o{e}^{-\left(\frac{2\pi mv}{{\mu }_{o}qI}\right)}$

Answer 1

The correct option is D $x=x_o{e}^{-\left(\frac{2\pi mv}{{\mu }_{o}qI}\right)}$


As there is no component of initial velocity along $Z-$direction, hence, the motion of the particle will take place only in the $XY-$ plane.

The force acting on the particle is,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=q(v_x\hat{i}+v_y\hat{j})\times \left(\frac{{\mu }_{o}I}{2\pi x}\hat{k}\right)$

$\therefore \overrightarrow{F}=q{v}_y\frac{{\mu }_{o}I}{2\pi x}\hat{i}-q{v}_x\frac{{\mu }_{o}I}{2\pi x}\hat{j}$

Now, acceleration along $X-$axis,

$a_x=\frac{F_x}{m}=\frac{{\mu }_{o}qI}{2\pi m}\frac{v_y}{x}=\lambda \frac{v_y}{x}...\left(i\right)$

Here, $\lambda =\frac{{\mu }_{o}qI}{2\pi m}=$ constant

As, $a_x=\frac{v_{x}d{v}_x}{dx}...\left(ii\right)$

Also, $v_x^2+v_y^2=v^2$

After differentiating with respect to time, we get,

$v_{x}d{v}_x=-v_{y}d{v}_y.....\left(iii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, we get,

$\lambda \frac{v_y}{x}=\frac{v_{x}d{v}_x}{dx}$

$\therefore v_{x}d{v}_x=\frac{\lambda v_{y}dx}{x}....\left(iv\right)$

From $\left(iii\right)$ and $\left(iv\right)$, we get,

$\frac{\lambda v_{y}dx}{x}=-v_{y}d{v}_y$

$\frac{dx}{x}=-\frac{d{v}_y}{\lambda }.....\left(v\right)$

Initially, $x=x_o$ and $v_y=0$.

Also, at minimum separation of the particle from the wire, $v_x=0$, so, $v_y=v$

Integrating equation $\left(v\right)$ using above limiting conditions, we get

${\int }_{x_o}^x\frac{dx}{x}={\int }_0^v-\frac{d{v}_y}{\lambda }$

$\Rightarrow \ln \frac{x}{x_o}=-\frac{v}{\lambda }$

Putting, $\lambda =\frac{{\mu }_{o}qI}{2\pi m}$

$\Rightarrow \ln \frac{x}{x_o}=-\frac{2\pi vm}{{\mu }_{o}qI}$

$\therefore x=x_o{e}^{-\left(\frac{2\pi mv}{{\mu }_{o}qI}\right)}$

Hence, option $\left(D\right)$ is the correct answer.