A square-shaped circuit with side a lies on the xy plane (the origin is in the center of the square and its sides are parallel to the axes). If a current of intensity I flows through the circuit, calculate the magnetic field B at a) the cent of the square, b) at the point (a / 2, a / 2), c) at the point (a, 0,0 ) and d) at the point (a / 2,0, -a / 2)

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Answer:-Let current flows clockwire in circuit as direction shown in figure. Magnetic field duc to wire forming angle \phi_{1}, \phi_{2} at some point is given byWe can tell dircction of field by right hand palm rulea) All four wires causes magnetic field going inside page, \phi_{1}=\phi_{2}=45^{\circ}, d=a / 2For all fowr wiresB=\frac{\mu_{0} I}{4 \pi d}\left(\sin \phi_{1}+\sin \phi_{2}\right) \otimesthe signshows field goes inide the pageso let \vec{B} due to A D=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left[\sin 45^{\circ}+45^{\circ}\right] \otimesso cotal \vec{B} due to all four wires =4 \vec{B}_{A D}=\frac{2 \mu_{0} I}{\pi a}\left[\frac{2}{\sqrt{B}}\right] \otimes \left.\vec{B}=\frac{2 \sqrt{2} \mu_{0} I}{\pi a}\right] \otimes{ }_{p} going inende page(b)We need to calculate field at c \equiv(a / 2, a / 2)Field duc to D C, B B=0as point lies on the wires D C, C BDue to AD\begin{array}{l}\vec{B}_{A D}=\frac{\mu_{0} I}{4 \pi a}\left(\sin 0^{\circ}+\sin 45^{\circ}\right) \otimes \\\vec{B}_{A B}=\vec{B}_{A D} \\\text { so } \vec{B}_{\left(\frac{a}{2}, \frac{a}{2}\right)}=\vec{B}_{A B}+\overrightarrow{B B}_{A D}=2 \vec{B}_{A B}=\frac{\mu_{0} I}{2 \sqrt{2} \pi a} \otimes\end{array}i^{0 \circ} 45by symmetry these two are equal\begin{array}{l}\vec{B}_{A D}=\frac{\mu_{0} I}{4 \pi\left(\frac{3 a}{2}\right)}[\sin \theta+\sin \theta]=\frac{\mu_{0} I}{3 \pi a}\left[\frac{1}{\sqrt{10}}\right] \cdot \theta \\\vec{B} B=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]=\frac{\mu_{0} I}{\sqrt{2} \pi a} \otimes \\\overrightarrow{B C D}=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left[\sin \left(-45^{\circ}\right)+\sin \left[90^{\circ}-\theta\right]\right]=\vec{B}_{A B} \\\text { so } \left.\vec{B}=\frac{\mu_{0} I\left[\frac { 1 } { \pi a } \left[\frac{1}{3 \sqrt{10}}+\frac{\sqrt{2}}{\sqrt{2}}+\left(\frac{-1}{\sqrt{2}}\right)+\cos 90^{\circ} \theta-\cos 90^{\circ} \sin \theta\right.\right.}{0}\right] \otimes \\\end{array}{ }_{B}(a, 0)=\frac{\mu_{0} I}{\pi a}\left[\frac{1}{3 \sqrt{10}}+\frac{3}{\sqrt{10}}\right]=\frac{\sqrt{10} \mu_{0} I}{3 \pi a} \otimesDrswer (C)(d) at (a / 2,0,-a / 2)P lies on x z planeFor A D\begin{aligned}d=E P & =\sqrt{\left[\frac{a}{2}-\left(-\frac{a}{2}\right)\right]^{2}+\left(0-\frac{a}{2}\right)^{2}} \\& =\frac{\sqrt{5} a}{2}\end{aligned}on\sin \theta_{1}=\sin \theta_{2}=\frac{D E}{E P}=\frac{a / 2}{\sqrt{\frac{5}{2} a}}=\frac{1}{\sqrt{5}}(1) shan disection of 2 anix\begin{array}{l}\mid \vec{B}_{A D}=\frac{\mu_{0} I}{4 \pi\left(\frac{\sqrt{5} a}{2}\right)}\left(\frac{2}{\sqrt{5}}\right)=\frac{\mu_{0} I}{5 \pi a} \hat{P X} \quad \hat{P X}=\left(\frac{-1}{\sqrt{5}} \hat{x}-\frac{-2}{\sqrt{5}} \hat{z}\right) \\\text { dareation } \vec{B}_{D C}=\frac{\mu_{0} I}{4 \pi(p C)}\left(\frac{D C}{D P}+\hat{\theta}\right) \hat{P} \hat{y} \quad \hat{P Y}=\left(\frac{1}{\sqrt{2}} \hat{y}-\frac{1}{\sqrt{2}} \hat{z}\right)\end{array}\begin{array}{l}\vec{B}_{D C}=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{\sqrt{2}}\right)}\left(\frac{a}{\sqrt{2 a \frac{5}{4} a^{2}+\frac{a^{2}}{4}}}\right)\left(\frac{1}{\sqrt{2}} \hat{y}-\frac{1}{\sqrt{2}} \hat{z}\right) \\=\frac{\mu_{0} I}{-\pi a}\left[\frac{\sqrt{2}}{A_{2}} \times \frac{2}{\sqrt{6}}\right]\left(\frac{\hat{y}}{\sqrt{2}}-\frac{\hat{z}}{\sqrt{2}}\right) \\=\frac{\mu_{0} I}{2 \sqrt{3} \pi a}\left(\frac{\hat{y}}{\sqrt{2}}-\frac{\hat{z}}{\sqrt{2}}\right) \\\end{array}\text { also } \begin{array}{l} \vec{B}_{A B}=\frac{\mu_{0} I}{2 \sqrt{3} \pi a}\left(-\frac{\hat{y}}{\sqrt{2}}-\frac{\hat{z}}{\sqrt{2}}\right) \vec{B}_{B C}=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \hat{n} \\\vec{B}_{B C}=\frac{\mu_{0} I}{\sqrt{2} \pi a} \hat{x}\end{array}\begin{array}{l}=\frac{\mu_{0} I}{5 \sqrt{5} a} \hat{x}-\frac{2 \mu_{0} I}{5 \sqrt{5} a}+\frac{\mu_{0} I}{2 \sqrt{3} \pi a}(-\sqrt{2} \hat{z})+\frac{\mu_{0} I}{\sqrt{2} \pi a} \hat{x} \\\left.=\sqrt{\frac{\mu_{0}}{\pi a}\left[\left(\frac{1}{\sqrt{2}}-\frac{1}{5 \sqrt{5}}\right) \hat{x}\right.}\left(\frac{1}{\sqrt{6}}+\frac{2}{5 \sqrt{5}}\right) \hat{z}\right] \\\end{array}Dswer (d) ...