Answer: 

Assume current in loop 1 as $I_1$ in clockwise direction . The current  $I_1$leaves the dotted terminal of $j5 \Omega$ while current  $I_2$ enters the dotted terminal of $j10 \Omega$, so we apply dot rule 2.

The controlled sources $j2I_2$ is connected in series with $j5$ and source $j2I_1$is connected in series with $j10$.

Applying KVL to the input loop , we get

$100\measuredangle0^0 -j5I_1+j2I_2-(3-j2)(I_1-I_2)=0$

$(3+j3)I_1-3I_2=100$...............(1)

Applying KVL to the output loop , we get 

$-j10I_2+j2I_1-5I_2-(3-j2)(I_2-I_1)=0$

$-3I_1+(8+j8)I_2=0$.................(2)

Writing the equation (1) and (2) in matrix form

$ \begin{bmatrix} 3+j3 & -3 \\[0.3em] -3 & 8+j8 \\[0.3em] \end{bmatrix} \begin{bmatrix} I_1 \\[0.3em] I_2 \\[0.3em] \end{bmatrix}= \begin{bmatrix} 100 \\[0.3em] 0 \\[0.3em] \end{bmatrix}$

Using cramers rule, we get

$I_2= \dfrac{\begin{bmatrix} 3+j3 & 100 \\[0.3em] -3 & 0 \\[0.3em] \end{bmatrix}} { \begin{bmatrix} 3+j3 & -3 \\[0.3em] -3 & 8+j8 \\[0.3em] \end{bmatrix}} $

$I_2=\dfrac{300}{-9+j48}$

$I_2=\dfrac{300}{48.84 \measuredangle+100.62}$

$\therefore I_2=6.143 \measuredangle -100.62 A$