A long, straight wire with a cross-sectional area A carries a...

The drift velocity of free electrons in the wire is given by $v_d = \dfrac{i}{nAe}$.

The direction of the drift velocity of electrons is opposite to the direction of the conventional current.

The trolley moves in the direction opposite to the current with speed $v = \dfrac{i}{nAe}$, which is exactly equal to the drift velocity of the electrons.

In the reference frame of the trolley, the free electrons are at rest, so they do not contribute to the current.

However, the positive ions of the metallic lattice, which are stationary in the ground frame, appear to move with velocity $v$ in the direction of the current.

The linear charge density of the positive ions is $\lambda_+ = nAe$.

The current due to the apparent motion of positive ions in the trolley frame is $i' = \lambda_+ v = (nAe) \left(\dfrac{i}{nAe}\right) = i$.

Since the net current in the trolley frame remains $i$, the magnetic field observed by the person is $B = \dfrac{\mu_0 i}{2\pi r}$.