1. For a conducting plate, the induced emf is given by Faraday’s law:

$|\mathcal{E}| = \left|\frac{d\Phi}{dt}\right|$

Since the magnetic field is normal to the plate,

$\Phi = BA$

So,

$|\mathcal{E}| = A\left|\frac{dB}{dt}\right|$

2. Given area of square plate:

$A = 4\,\text{m}^2$

3. From the graph, during the interval $t=2\,\text{s}$ to $t=4\,\text{s}$, the magnetic field changes linearly by $4\,\text{mT}$ in $2\,\text{s}$.

Hence,

$\left|\frac{dB}{dt}\right| = \frac{4\times 10^{-3}}{2} = 2\times 10^{-3}\,\text{T/s}$

4. Therefore,

$|\mathcal{E}| = 4 \times 2\times 10^{-3} = 8\times 10^{-3}\,\text{V}$

$|\mathcal{E}| = 8\,\text{mV}$

5. Final answer:

$\boxed{8}$

Comparison with stored answer: The derived answer is $8$, which matches the stored correct answer.