In closed circuit shown in Fig., $AB, BC$ and $CD$ are straight conductors, each of length $R$ and $DEA$ is a semicircle of radius $R$. A small circular loop of radius $r$ is coplanar with the circuit and center of loop coincides with center of curvature of the semicircle. If current through the circuit increases at a constant rate $\frac{dI}{dt}=\alpha$. Calculate emf induced in the loop. 

EMF induced in the loop is rate of increase of flux linked with it. To calculate it, magnetic induction at center of loop, as function of current $I$ flowing through the circuit, is to be calculated. 

Since part $DEA$ of the circuit is semicircle of radius $R$, therefore, $AD$ is diameter and its length is $2R$ and length of each of three straight conductors $AB, BC, CD$ is $R$, hence, these straight conductors from a part of regular hexagon.

Due to semicircle part, magnetic induction at center is 

$B_1=\frac{{\mu }_{0}nI}{2R}=\frac{{\mu }_0\left(\frac{1}{2}\right)I}{2R}$ (inward)

Now considering one straight conductor as shown in Fig, 

$\alpha =\beta =30° ; r=R\cos 30°$

$B_2=\frac{{\mu }_{0}I}{4\pi r}(\sin \alpha +\sin \beta )=\frac{\sqrt{3}{\mu }_{0}I}{6\pi R}$ (inward)

Therefore, Resultant magnetic induction at center is 

$B=B_1+3B_2=\frac{{\mu }_{0}I}{4\pi r}(\pi +2\sqrt{3})$

Area of the loop $=\pi r^2$

Therefore, flux linked with it, $\varphi =\pi r^{2}B=\frac{{\mu }_0(\pi +2\sqrt{3})r^2}{4R}I$ 

Magnitude of induced emf $=\left|\frac{d\varphi }{dt}\right|=\frac{d\varphi }{dI}.\frac{dI}{dt}=\frac{{\mu }_0(\pi +2\sqrt{3})r^2\alpha }{4R}$ Ans.