A structure formed from two straight conducting rods forming a right angle, is placed in a uniform magnetic field with B = 0.40 T is directed out of the page. A straight 
conducting bar in contact with the structure, starts at the vertex at time t = 0 and moves with a constant velocity of 5.0 m/s along them. The emf (in V) around the triangle at
that time t=4.0 s is __________.

The perpendicular length of the triangular area enclosed by the structure and bar is the same as the distance travelled by it in time t

d=vt, where v=5.0 m/s.

Here the "base" of that triangle (the distance between the intersection points of the bar with the structure) is 2d. Thus, the area of the triangle is

$A=\frac{1}{2}\left(\text{ base }\right)\left(\text{ height }\right)=\frac{1}{2}\left(2vt\right)\left(vt\right)=v^2{t}^2$

Since the field is a uniform B = 0.350 T, then the magnitude of the flux (in SI units) is

${\varphi }_B=BA=\left(0.40\right)(5.0{)}^2{t}^2=10.0t^2$

The magnitude of the emf is the (absolute value of) Faraday's law:

$\begin{array}{l}\epsilon =\frac{d{\varphi }_B}{dt}=10.0\frac{d{t}^2}{dt}=20.0t\\ \text{ At }t=4.0\mathrm{s},\text{ this yields }\epsilon =80.0\mathrm{V}\end{array}$