1. Magnetic field due to the long straight wire

At a distance $r$ from a long straight wire carrying current $I$,

$B(r)=\frac{\mu_0 I}{2\pi r}$

The particle is released moving parallel to the wire, so initially its velocity is along the current direction.

2. Direction of magnetic force

Magnetic force on the particle is

$\vec F = q\,\vec v \times \vec B$

Since $\vec v$ is along the wire and $\vec B$ is tangential around the wire, the force is radially inward (towards the wire), so the particle moves closer to the wire.

Also, because magnetic force is always perpendicular to velocity,

$|\vec v| = v_0 = \text{constant}$

So the speed remains unchanged.

3. Choose cylindrical coordinates

Let the wire be along the $z$-axis. Then the particle moves in the $rz$-plane with:

• radial coordinate $r$
• axial coordinate $z$

Since the force has no $z$-component, the $z$-component of momentum remains constant:

$M v_z = \text{constant} = M v_0$

So,

$v_z = v_0$

for all time.

But total speed is constant and equal to $v_0$:

$v^2 = v_r^2 + v_z^2 = v_0^2$

This would seem to give $v_r=0$, which is impossible if the particle moves inward. Hence we must be careful: as the particle moves in cylindrical geometry, there is also azimuthal motion generated by the magnetic force. So we should use a conserved quantity approach.

4. Use vector potential / conserved canonical momentum

For the magnetic field of a long straight wire,

$B_\phi = \frac{\mu_0 I}{2\pi r}$

A convenient vector potential is

$A_z = -\frac{\mu_0 I}{2\pi}\ln r$

Since the system is independent of $z$, the canonical momentum along $z$ is conserved:

$p_z + qA_z = \text{constant}$

Initially, at $r=a$, the particle has only $z$-velocity $v_0$, so

$M v_0 + q\left(-\frac{\mu_0 I}{2\pi}\ln a\right)=\text{constant}$

At a general distance $r$,

$M v_z + q\left(-\frac{\mu_0 I}{2\pi}\ln r\right)= M v_0 - \frac{q\mu_0 I}{2\pi}\ln a$

Thus,

$M v_z = M v_0 + \frac{q\mu_0 I}{2\pi}\ln\frac{r}{a}$

So,

$v_z = v_0 + \frac{q\mu_0 I}{2\pi M}\ln\frac{r}{a}$

5. Condition at the turning point

At the minimum distance $r=x$, the radial velocity becomes zero. Since magnetic force does no work, total speed remains $v_0$.

At the turning point, radial speed is zero, so all the speed is along the $z$-direction:

$|v_z| = v_0$

Initially $v_z=+v_0$. As the particle is attracted and turns around radially, at the closest approach it must momentarily have

$v_z=-v_0$

So put $r=x$ and $v_z=-v_0$ in the conservation equation:

$-M v_0 = M v_0 + \frac{q\mu_0 I}{2\pi}\ln\frac{x}{a}$

Therefore,

$-2Mv_0 = \frac{q\mu_0 I}{2\pi}\ln\frac{x}{a}$

$\ln\frac{x}{a} = -\frac{4\pi M v_0}{q\mu_0 I}$

Hence,

$\frac{x}{a}=e^{-\frac{4\pi M v_0}{q\mu_0 I}}$

So,

$\boxed{x=a\,e^{-\frac{4\pi M v_0}{q\mu_0 I}}}$

6. Compare with options

This matches Option B:

$\boxed{a e^{-\frac{4 \pi m v_0}{q \mu_0 I}}}$

(Here $m$ and $M$ both denote the mass of the particle.)

7. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So they agree.