Physics - Magnetism Question with Solution | TestHub

$\frac{\mathit{\text{x}}}{\mathit{\text{d}}}=\text{sin}45° \Rightarrow \mathit{\text{x}}=\frac{\mathit{\text{d}}}{\sqrt{2}}$
we can extend the wire and then subtract the field of the extended part.

${\mathit{\text{B}}}_1=\frac{{\mu }_0}{4\pi }\frac{\mathit{\text{I}}}{\mathit{\text{d}}/\sqrt{2}}\left(\text{sin}0°+\text{sin}90°\right)-\frac{{\mu }_0}{4\pi }\frac{\mathit{\text{I}}}{\mathit{\text{d}}/\sqrt{2}}\left(\text{sin}0°+\text{sin}45°\right)$

$=\frac{{\mu }_0}{4\pi }\frac{\sqrt{2}\mathit{\text{I}}}{\mathit{\text{d}}}\left(1-\frac{1}{\sqrt{2}}\right)=\frac{{\mu }_0\mathit{\text{I}}}{2\sqrt{2}\mathrm{ \pi }\mathit{\text{d}}}\left(1-\frac{1}{\sqrt{2}}\right)\otimes$

$B_1\otimes \text{,}{\mathit{\text{B}}}_2\otimes \Rightarrow B=\left(B_1+B_2\right)\oplus =2B_1\otimes =\frac{{\mu }_{0}I}{\sqrt{\text{2}}\mathrm{ \pi }d}\left(1-\frac{1}{\sqrt{2}}\right)\otimes$