If a wire loop moves at constant speed into a region where there is a magnetic field, why is a current induced in the wire?

In this question, a square wire loop of length L and resistance R starts in a region with no magnetic field, and moves at a speed v into a region with magnetic field B, which is acting perpendicular to the area of the loop. The two regions are separated by a line parallel to the front edge of the loop. When the square loop is moving in the region with no magnetic field, there is no flux through the loop. However, once the front side of the loop reaches the region where there is a magnetic field, the flux through the loop will start to increase. If we set the moment when the front side of the loop reaches the boundary between the two regions as time t=0, then at time t after that, the area of the loop through which the magnetic field can pass is A=L*vt. The flux is given by the equation phi=BA, so the flux through the loop is phi=BLvt.After time t=L/v, the back edge of the loop will reach the boundary between the 2 regions, and so at any time after that the area is the maximum, L^2. Therefore the flux will remain constant at phi=BL^2. The emf, which is equivalent to the voltage, is related to the flux by the equation emf = d/dt(phi), i.e. the emf is the time derivative of the flux. The time derivative of BLvt is BLv, so between times t=0 and t=L/v, the emf is constant at BLv. Before time t=0, the flux is zero, so the emf is zero. After time t=L/v, the flux is constant, so the time derivative, i.e. the emf, is zero. The current is related to the emf by the equation V=IR, where V is the emf and I is the current, so the current induced in the loop is BLv/R between t=0 and t=L/v, and zero otherwise.