Photoelectric Effect (and Electric Current)

This question and its answer may help explain the relationship between intensity and number of photons. As Thomas Fritsch pointed out, electric current is the number of electrons emitted per unit time. Assuming the photons' frequency and hence energy is high enough to knock an electron loose, more incident photons per unit time means more photoelectrons per unit time, which is a higher current.

EDIT: The second answer to the linked question is helpful for your comment. From the first answer we know that intensity is proportional to the product of number of photons and photon energy: $$Ipropto nnu$$ where $I$ is intensity, $n$ is number, and $nu$ is frequency. From the second we know that intensity is proportional to the amplitude squared: $$Ipropto E^2$$ where $I$ is intensity and $E$ is amplitude (I use E here because the first answer in the other question already used A for area.) Therefore $$E^2propto nnu$$ That shows how to relate amplitude to the other parameters.

Increasing frequency does increase intensity, but it doesn't increase $n$, the number of photons, so it doesn't increase the number of photoelectrons and hence doesn't increase the current.

EDIT 2: Above I gave the relationship between intensity and the amplitude of a classical wave. Intensity $I$ is defined as power per unit area, and power $P$ is rate of energy flow. Each photon has energy $hbar nu$, so if the rate at which photons land on your detector is $R$ and the area of your detector is $A$, then $$I=frac{P}{A}=frac{Rhbarnu}{A}$$