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S Sog \rightarrow Matic field due to ciscular doop =\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}R \rightarrow Radius of loopI \rightarrow Grrent in loopx \rightarrow Distance of point fromcentre of loopMagnetic field due to long straight wire =\frac{\pi 0}{2 \pi} \frac{I}{r}I \rightarrow current in wirer \rightarrow distance of point from wire\therefore Magnetic field. due to straight wire =\frac{\mu_{0}}{2 \pi} \times \frac{\pi}{3 \pi}=\frac{2 \times 10^{-4} \times 10}{0.5}=4 \mu TThis feld acts in-ve z direction.(from Right hand thumb Rule, Keeping thum in tve ydirection. fingers points in -v e z diraction at point P )Magnetic field due to doop \rightarrow \frac{40}{2} \frac{I R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}\begin{array}{l}=\frac{2 \pi \times 10^{-7} \times 9 \times(0.5)^{2}}{\left(0.5^{9}+0.5^{2}\right)^{3 / 2}} \\=\frac{2 \pi \times 10^{-7} \times 9 \times 0.25}{(0.25+0.25)^{3 / 2}} \\=39.98 \times 10^{-7} \\=3.998 \times 10^{-6} \\=4 \mu T \text { (along-vex-axis) }\end{array}(curling fingers along direction of current perpendicular. thumb gives disuction of magnetic field, ie along-ve x-axis\therefore \vec{B}=(-4, T) \hat{\imath}-(4, \pi T) \hat{k}option (D) ...