electromagnetism - How to calculate induced emf of moving rod?

I have the following exercise:

A copper rod with length L is moving on two frictionless conducting rails, in a distance a from an very long straight conductor. The rods resistance: R. Ignore any self inductance. The magnetic permeability is $\mu_0$

a) Find the direction and magnitude of the induced current.

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I know I could probably just use faradays law finding the induced emf:
$\mathcal{E}=-\frac{d \Phi_{B}}{d t}$

But instead I tried using this other version, and I just can't get it working: $$ \begin{array}{l}{\mathcal{E}=\oint(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \cdot d \vec{l}} \\ {\text { (all or part of a closed loop moves }} \\ {\text { in a } \overrightarrow{\boldsymbol{B}} \text { field) }}\end{array} $$ I think I'm a bit confused about the direction of the dL-vector

Here is my attempt: Let say the dl-vector is along the x-axis and we integrate from a to a+L. The B-field is into the plane, and the $$ \mathcal{E}=\int_{a}^{a+L}(\vec{v} \times \vec{B}(L)) \cdot d \vec{L} =\int_{a}^{a+L}(v \hat{k} \times B(L) \hat{\jmath}) \cdot d L \hat{i}=\int_{a}^{a+L}-v \cdot B(L) d L$$ $$=\int_{a}^{a+L}-v \cdot \frac{\mu_{0} I}{2 \pi L} d L=-\frac{v \mu_{0} I}{2 \pi}(\ln (a+L)-\ln (a)) $$ $$=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a}{a+L}\right)$$

I know this is the wrong emf, I'm supposed to get $$\mathcal{E}=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{L}\right)$$

And even if I chose my dL-vector to go in the other direction and integrate from a+L to a. I would get the same result.

What am I doing wrong here?

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Edit: So the reason why I thought my result was wrong, was since the answer of the induced current was: $$ I=\frac{v \mu_{0} I}{2 \pi R} \ln \left(\frac{a}{a+L}\right) $$

But I think it comes from faradays law, where you have a negative sign in front of the induced emf. When you want to find the current you should take the magnitude. So I shouldn't multiply with the negative sign and change the fraction in the natural log. I should leave negative sign in front.