a b c d

b

$\text{EMF in the left part of wire is }e_1 = \frac{\mu_0 iav}{2\pi(x -\frac{a}{2})}$ $\text{EMF in the Right part of wire is }e_2 = \frac{\mu_0 iav}{2\pi(x +\frac{a}{2})}$ $ \text{Net Emf } e = e_1 - e_2 = \frac{\mu_0 iav}{2\pi (x -\frac{a}{2})} -  \frac{\mu_0 iav}{2\pi(x +\frac{a}{2})} = 1\mu V$