A square non-conducting loop, 20 cm, on a side is placed in a magnetic field. The centre of side AB coincides with the centre of magnetic field. The magnetic field is increasing at the rate of 2 T/s. The potential difference between B and C is

Perpendicular distance between BC and center O is d.
 

From B , select a small element dx at a distance of x.

Induced electric field is tangential to circle of radius $\sqrt{d^2+x^2}$.

$$\begin{gathered} \int \overrightarrow{E}.\overrightarrow{dl}=\left|\frac{d\varphi }{dt}\right|=A\frac{dB}{dt} \\ \Rightarrow E\left(2\pi r\right)=\left(\pi {\left(r\right)}^2\right)\frac{dB}{dt} \\ \Rightarrow E=\frac{r}{2}\frac{dB}{dt} \left[here r=\sqrt{d^2+x^2}\right] \\ Now, from B to C \\ △V=\int \left(E\cos \theta \right)dx \\ \Rightarrow △V=\int \frac{r}{2}\frac{dB}{dt}\times \frac{d}{r}\times dx \\ \Rightarrow △V=\frac{d}{2}\frac{dB}{dt}\left({\mathcal{l}}_{BC}\right)=\left(\frac{0.10}{2}\right)\left(2\right)\left(0.2\right)=0.02V=20mV \end{gathered}$$