Set up the geometry

Let the two conducting rails make an angle $\theta$ at the vertex.
The conducting bar always connects the two rails and moves away from the vertex with constant speed $v$.

If the distance of the bar from the vertex along the angle bisector (or equivalently, the characteristic linear distance from the vertex) is $x$, then since the bar moves with constant velocity, $x = vt$ so $x \propto t.$

Length of the rod as a function of position

The separation between the rails increases linearly with distance from the vertex. Therefore, the length of the rod between the rails is $\ell \propto x.$ Since $x \propto t$, we get $\ell \propto t.$

Area enclosed by the circuit

The loop formed by the two rails and the rod is a triangular sector-like region. Its area is proportional to $A \propto (\text{base})(\text{height}) \propto x^2.$ More explicitly, for rails inclined at fixed angle $\theta$, $A = \frac{1}{2} \ell x \propto x^2.$ Since $x = vt$, $A \propto t^2.$

Magnetic flux

Magnetic field $B$ is constant and perpendicular to the plane, so the flux is $\Phi = BA \propto t^2.$

Induced EMF

By Faraday’s law, $\mathcal{E} = \left|\frac{d\Phi}{dt}\right|.$ Since $\Phi \propto t^2$, $\mathcal{E} \propto t.$

Hence, comparing with $\mathcal{E} \propto t^n,$ we get $n = 1.$

Comparison with stored answer

Derived answer: $1$
Stored correct answer: $1$

They match.