Answer
See the detailed answer below.
Work Step by Step
Let's assume that the current carrying wire is too long at which the magnetic field exerted by it is given by
$$B=\dfrac{\mu_0I}{2\pi x}\tag 1$$
Noting that the magnetic field direction due to the long wire is into the page while the short wire is moving upward. According to the right-hand rule, the magnetic force on the positive charges will be to the left.
This means that the positive charges will build up on the left side of the short metal wire while the negative charges will build up on the right side of the wire.
This creates a building electric field until the electric force due to this
electric field equals the magnetic force.
Thus,
$$F_E=F_B$$
$$ \color{red}{\bf\not} qE= \color{red}{\bf\not} qvB$$
$$ E= vB$$
Plug from (1),
$$E=\dfrac{\mu_0vI}{2\pi x} \tag 2$$
Recalling that
$$\Delta V=-\int Eds$$
So, in this case, integrating the electric field from the positive end of the wire (which is the left end) to its negative end (which is the right end).
$$\Delta V= \int_d^{d+l} Edx$$
Plugging from (2),
$$\Delta V= \int_d^{d+l} \dfrac{\mu_0vI}{2\pi x} dx$$
$$\Delta V= \dfrac{\mu_0vI}{2\pi }\int_d^{d+l} \dfrac{ dx}{x}$$
$$\Delta V= \dfrac{\mu_0vI}{2\pi } \ln(x)\bigg|_d^{d+l}$$
$$\boxed{\Delta V= \dfrac{\mu_0vI}{2\pi } \ln\left(\dfrac{d+l}{d}\right) }$$
