An example regarding Faraday's and Lenz's Law [closed]

Consider the following configuration :

$$\hskip2.48in$$

Here we have a closed loop near a current carrying wire with the given direction. If we need to calculate the induced emf which results because of moving the loop upward with a constant speed $$v$$, then we need to calculate the flux $$\Phi$$ penetrating the area of this loop as: $$\int B \cdot dS = \cfrac{\mu_0 I}{2\pi} \int^{s+a}_s \cfrac{1}{x} dx = \cfrac{\mu_0 I}{2\pi} ln(\cfrac{s+a}{s})$$

Now, if we need to calculate the induced emf, we will do the following $$\mathcal{E} = -\cfrac{d\Phi}{dt}=-\cfrac{\mu_0 I a}{2 \pi}\cfrac{d}{dt}ln(\cfrac{s+a}{s})$$

Now my first question arises here, more specifically in the differentiation step of the $$ln(x)$$ function. Did the result of the differentiation arisen by assuming that s is a function of time because the loop moves upward with a constant $$v$$ and therefore the distance $$s$$ changes with time? As $$s = f(t)$$ $$\cfrac{d}{dt} [ln(\cfrac{s+a}{s}) ] = \cfrac{d}{dt}[ln(s+a) - ln(s)] = \cfrac{1}{s+a} s' - \cfrac{1}{s} s'$$

Is this is right ?

Finally , I need to test my understanding of Lenz's law by finding the direction of the induced current. Here the field is out of the screen so to oppose the change of this field we need to the induced field to be in the screen and therefore to achieve this direction of field the current in the loop must be in the clockwise direction and this will be the direction of the induced current. Is this right ?

closed as off-topic by Aaron Stevens, Jon Custer, Kyle Kanos, Dvij Mankad, Rory AlsopMay 20 at 15:16

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1. Yes, it is perfectly valid. Here, $$\cfrac{d^2 \Phi}{dt^2} \ne 0$$ ,i.e. The manner in which the magnetic flux through the loop changes is not uniform with respect to time, which implies that the change of change in magnetic field is dependent on the position of loop with respect to the wire, so the $$\mathcal{E}$$ expression you have written is the induced emf of the particular instant, unlike the case when $$\cfrac{d^2 \Phi}{dt^2} = 0$$, where $$\mathcal{E}$$ is independent of distance.