Find Voltage induction flux through the circuit
The approximation you used can not be applied that way because it hides a very special expansion. Please consider the algebraic manipulation:

$$ \ln \left (\frac{d+\frac{a}{2}}{d-\frac{a}{2}}  \right )= \ln \left(1+\frac{a}{d-\frac{a}{2}} \right ) $$

Now, the well-known expansion:
$$ \ln \left (1+x \right ) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots $$
In our case:

$$ \ln \left(1+\frac{a}{d-\frac{a}{2}} \right ) = \frac{a}{d-\frac{a}{2}} - \frac{1}{2} \left (\frac{a}{d-\frac{a}{2}} \right )^2 + \frac{1}{3} \left (\frac{a}{d-\frac{a}{2}} \right )^3 - \frac{1}{4} \left (\frac{a}{d-\frac{a}{2}} \right )^4 + \cdots $$

If \$ a \ll d \$, we can do the first term approximation:

$$ \ln \left(1+\frac{a}{d-\frac{a}{2}} \right ) \approx  \frac{a}{d-\frac{a}{2}} $$

Or further:
$$ \ln \left(1+\frac{a}{d-\frac{a}{2}} \right ) \approx  \frac{a}{d} $$

Putting this result on expression for induced flux:
$$ \phi = \frac{\mu I}{2\pi}\cos \omega t \ln \left ( \frac{d+\frac{a}{2}}{d-\frac{a}{2}} \right )a $$
$$ \phi = \frac{\mu Ia^2}{2\pi d}\cos \omega t  $$

The corresponding induced voltage:

$$ V = -\frac{\mathrm{d} \ }{\mathrm{d} t}\phi(t) $$
or
$$ V = \frac{\mu \omega a^2 I}{2\pi d}\sin \omega t $$