A rectangular frame $ABCD$, made of a uniform metal wire has a straight connection between $E$ and $F$ made of the same wire as shown in the figure. $AEFD$ is a square of side $1m$ and $EB=FC=0.5m$. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is $1T/s$. The resistance per unit length is $1\Omega /m$. Find the magnitude and directions of the current in segments $AE, BE$ and $EF$. 

Induced emf in loops $AEFD$ and $EBCF$ would be

$$\begin{gathered} e=\left|\frac{d{\varphi }_1}{dt}\right|=S_1\left|\frac{dB}{dt}\right| \\ =\left(1\times 1\right)=1V \end{gathered}$$

Similarly,  $$\begin{gathered} e_2=\left|\frac{d{\varphi }_2}{dt}\right|=S_2\left|\frac{dB}{dt}\right| \\ =\left(0.5\times 1\right)\times 1V \\ =0.5V \end{gathered}$$

Now, current is increasing, induced current will produce the magnetic field in direction, hence $e_1$ and $e_2$ will be applied as shown. 

By Kirchoff's first law at junction $F$, we get

$i_1=i+i_2$ …………………… (i)

Kirchoff's second law is loop $FEADF$

$3i_1+i=1$ …………………… (ii)

Kirchoff's second law is loop $FEBCF$

$2i_2-i=0.5$ ……………………. (iii)

On solving Eqs. (i), (ii) and (iii), we get

$$\begin{gathered} i_1=\left(\frac{7}{22}\right)A \\ {i}_2=\left(\frac{6}{22}\right)A \\ i=\left(\frac{1}{22}\right)A \end{gathered}$$

Therefore, current in segment $AE$ is $\left(\frac{7}{22}\right)A$ from $E$ to $A$, current in segment $BE$ is $\left(\frac{6}{22}\right)A$ from $B$ to $E$ and current in segment $EF$ is $\left(\frac{1}{22}\right)A$ from $F$ to $E$.