A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is:

We start with the magnetic (Lorentz) force formula

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

where $$q$$ is the charge, $$\vec v$$ its velocity and $$\vec B$$ the magnetic field produced by the straight current-carrying wire.

Let us choose a convenient co-ordinate system. Take the wire to lie along the $$z$$-axis and let the current $$I$$ flow in the $$+z$$ direction. The magnetic field produced by a long straight wire is given by the Biot-Savart law,

$$\vec B = \frac{\mu_0 I}{2\pi r}\,\hat\phi,$$

where $$r$$ is the perpendicular distance from the wire and $$\hat\phi$$ is the azimuthal unit vector obtained from the right-hand rule (curl the fingers of the right hand around the wire in the direction of the current; the thumb points along $$+z$$).

Pick a point lying on the positive $$x$$-axis at a distance $$r$$ from the wire. At this point the magnetic field is tangential and points along $$+y$$. Symbolically,

$$\vec B = B\,\hat y,\qquad B = \frac{\mu_0 I}{2\pi r}.$$

The test charge is negative. Suppose its velocity has only a radial component, either directly away from the wire (along $$+\hat x$$) or towards the wire (along $$-\hat x$$). We now calculate the cross product in each case.

If the charge moves away from the wire, we have

$$\vec v = v\,\hat x,$$

so

$$\vec v \times \vec B = \bigl(v\,\hat x\bigr) \times \bigl(B\,\hat y\bigr) = vB\,(\hat x \times \hat y) = vB\,\hat z.$$

Therefore the force becomes

$$\vec F = q\,\vec v \times \vec B = q\,vB\,\hat z.$$

Because $$q<0$$ (negative charge), the force points along $$-\,\hat z$$, i.e. opposite to the direction of current. This does not match the given condition that the force is parallel to the current.

If the charge moves towards the wire, we have

$$\vec v = -\,v\,\hat x,$$

then

$$\vec v \times \vec B = \bigl(-v\,\hat x\bigr) \times \bigl(B\,\hat y\bigr) = -vB\,(\hat x \times \hat y) = -vB\,\hat z.$$

The force is now

$$\vec F = q\,\vec v \times \vec B = q\,(-vB\,\hat z).$$

With $$q<0$$, the double negative gives

$$\vec F = (+vB)\,\hat z,$$

so $$\vec F$$ is indeed along $$+\,\hat z$$, the same direction as the current. This satisfies the condition stated in the problem.

Thus, for the force on a negative charge to be parallel to the current, the charge must be moving radially inward, i.e. it must be approaching the wire.

Hence, the correct answer is Option B.