A Conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the ${90}^0$ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 A$s^{-1}$ . Which of the following statement (s) is (are) true?

$\begin{array}{l}Flux=\phi =\int \overrightarrow{B}\cdot d\overrightarrow{A}={\int }_0^{0.1} 2xdx\frac{{\mu }_{0}i}{2\pi x}=\left(\frac{{\mu }_{0}i}{\pi }0.1\right)=\frac{{\mu }_{0}i}{10\pi }\\ \text{ Also, }\phi =Mi\\ \Rightarrow \text{ Mutual Inductance }=M=\frac{\phi }{i}=\frac{{\mu }_0}{10\pi }\\ \text{ Induced emf }=\epsilon =M\frac{di}{dt}=\left(\frac{{\mu }_0}{10\pi }\right)\left(10\right)=\frac{{\mu }_0}{\pi }\end{array}$

As the current in loop is increased, its flux is trying to increases out of the page. To oppose it, direction of current in wire should be towards right. Since, direction of current in wire is same as direction of current in hypotenuse , force b/w wire & loop is repulsive.
As the loop rotates  around the wire flux through it due to current in wire will not change. Hence M will not change. So, no additional emf is generated.