I am preparing for an exam, on one problem my answer differ from the solution.
The current $i(t)=I_0e^{-\alpha t}$ runs in a long straight conductor along $\hat z$.
Point A,B,C,D forms a triangle. A and B is at $L \hat x$ with a very small separation in $\hat z$ with a resistance R between. C and D is at $2L \hat x$ with a separation in $\hat z$ of $2L$.
The question is: What is the induced current in the triangle.
I calculated the flux $\Omega$ correctly, but my answer differ on the direction of the current.
I took $V_{ind}=-\frac{d\Omega}{dt}$, since $i(t)$ is decreasing, the magnetic flux passing through ABCDA is decreasing, and so (to my understanding) it is reasonable that $i_{ABCD}=\frac{V_{ind}}{R}$ will run clockwise (using right-hand notation).
However, in the solution to the problem they too use $V_{ind}=-\frac{d\Omega}{dt}$ but then "magically" on the next line remove the minus sign and says that according to Lenz law the current opposes the magnetic flux that caused the current and so will run counter-clockwise.
I would be thankful for an intuitive understanding for why it is opposing the magnetic flux that caused the current and not opposing the change in the magnetic flux? Thanks.