Question

A charge (q, m) is projected perpendicular away from an infinitely long wire carrying a current i with a speed $v_0$. Find the maximum separation between the charge and the wire in the subsequent motion if the initial separation is $x_0$.

• A. $3x_0$
• B. $x_0{e}^{\frac{\pi m{v}_0}{{\mu }_{0}iq}}$
• C. $x_0{e}^{\frac{2\pi m{v}_0}{{\mu }_{0}iq}}$
• D. $2x_0$

Answer 1

The correct option is C $x_0{e}^{\frac{2\pi m{v}_0}{{\mu }_{0}iq}}$

The situation from the question is as follows:
Let the velocity and acceleration of the charge be $\overrightarrow{v}$ and $\overrightarrow{a}$ respectively and the magnetic field be $\overrightarrow{B}$.
The force F on the charge = $q\overrightarrow{v}\times \overrightarrow{B}$ = $q(v_x\overrightarrow{i}+v_y\overrightarrow{j})\times \frac{{\mu }_{0}i}{2\pi x}(-\overrightarrow{k})$ = m ($a_x\overrightarrow{i}+a_y\overrightarrow{j})$
So,
$a_x=\frac{-q{v}_y{\mu }_{0}i}{2\pi xm}$

$a_y=\frac{q{v}_x{\mu }_{0}i}{2\pi xm}=\frac{d{v}_y}{dt}$.$\frac{dx}{dx}$ = $v_x\frac{d{v}_y}{dx}$

$\Rightarrow \frac{q{\mu }_{0}i}{2\pi xm}=\frac{d{v}_y}{dx}$

Now as the magentic field is always normal to the velocity of the charge, the speed of the charge is always $v_0$.
At the instant of maximum seperation between the charge and wire, $v_x=0$ such that $v_y=v_0$ directed in the positive y direction and $x=x_1$; and initially $v_x=v_0$ directed towrads the wire, $v_y=0$ and $x=x_0$.
So,
${\int }_{x_0}^{x_1}\frac{q{\mu }_{0}i}{2\pi xm}.dx={\int }_0^{v_0}d{v}_y$

Solving the above integral equation we get:
$x_1=x_0{e}^{\frac{2\pi m{v}_0}{{\mu }_{0}iq}}$

Hence, option (c) is correct.