Three identical wires are bent into circular arcs each of radius $r$ such that each arc subtends an angle $\theta =120°$ at its center of curvature. These arcs are connected with each other to form a closed mesh such that one of them lies in $x-y$ plane, one in $y-z$ plane and the other in $z-x$ plane as shown figure. In the region of space a uniform magnetic field of induction $\overrightarrow{B}=B_0(\hat{i}+\hat{j})$ exists, whose magnitude increases at a constant rate $\frac{dB}{dt}=\alpha$. Calculate magnitude of emf induced in the mesh and mark direction of flow of induced current in the mesh shown in Fig.

To calculate magnitude of induced emf, magnetic flux linked with the mesh must be known. Magnetic induction $\overrightarrow{B}$ has two components of same magnitude $B_0$, one along positive $x-$direction and the other along positive $y-$direction.

To calculate flux linked with the mesh due to $x-$component of magnetic field, system must be viewed normal to $y-z$ plane or along $x-$axis. Then it appears as shown in Fig. (a). In this figure, positive $x-$direction is outward normal to plane of this paper. Hence point $R$ overlaps with origin $O$.

Center of curvature of arc $PQ$ does not coincide with origin but it lies at $C$ as shown in Fig. (b)

Considering $∆OCQ$ as shown in Fig (b) according to sine rule, 

$\frac{OQ}{\sin 120}=\frac{r}{\sin 45}$

$\therefore OQ=r\sqrt{\frac{3}{2}}$

Area of this triangle,

$A_1=\frac{1}{2}\times \left(OQ\right)\times r\sin 15°=\frac{\sqrt{3}(\sqrt{3}-1)}{8}{r}^2$

Area subtended by the arc at center $C$,

$A_2=\frac{1}{3}\pi r^2$

Therefore, the total area of mesh appeared in the Fig. (a) 

$A=2A_1+A_2=\frac{3\sqrt{3}(\sqrt{3}-1)+4\pi }{12}{r}^2$

Flux linked with the mesh due to $x-$component of magnetic induction,

${\varphi }_1=\text{that component of magnetic induction}\times \text{area of mesh appearing in the figure}$

$\therefore {\varphi }_1=\frac{3\sqrt{3}(\sqrt{3}-1)+4\pi }{12}{B}_0{r}^2$

Similarly it can be proved that an equal amount of flux is linked with the mesh due to $y-$component of magnetic induction. 

Hence total flux linked is $\varphi =2{\varphi }_1$

                                           $=\frac{3\sqrt{3}(\sqrt{3}-1)+4\pi }{6}{B}_0{r}^2$

Therefore, magnitude of induced emf, $e=\frac{d\varphi }{dt}=\frac{3\sqrt{3}(\sqrt{3}-1)+4\pi }{6}{r}^2.\frac{d{B}_0}{dt}$ ………….(1)

But $\overrightarrow{B}=B_0(\hat{i}+\hat{j})$ therefore its magnitude is $B=B_0\sqrt{2}$

$\therefore \frac{dB}{dt}=\frac{d{B}_0}{dt}.\sqrt{2}=\alpha$

Hence, $\frac{d{B}_0}{dt}=\frac{\alpha }{\sqrt{2}}$.

Substituting this value in equation (1)

$e=\frac{3\sqrt{3}(\sqrt{3}-1)+4\pi }{6\sqrt{2}}\alpha r^2$ Ans.

Since, strength of magnetic field is increasing; therefore, flux linked is also increasing. Hence according to Lenz's law induced current should oppose increase of flux linked. Therefore its direction will be as shown in Fig. (c)