91Ó°ÊÓ

The current through wire is,$i=10 \text{A}$

The length of wire is,$a=10 \text{cm}$

For this problem, we will use the equation for magnetic field due to finite wire of length L, carrying currenti, at a point P above the right end of the wire, and at distanceR from the wire. Direction of the magnetic field will depend on the direction of the current;we can determine this by right hand thumb rule.

Formula:

Magnetic field due to finite wire of length L, carrying currenti, at a point P above the right end of the wire, and at distance R from the wire is

$B_p=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}R}·\left(\frac{L}{\sqrt{L^2+R^2}}\right)$

We break the wire loop in eight parts as shown in the figure.We can find the magnetic field due to each part at point P

Due to part 1 magnetic field is $B_1$

$$\begin{gathered} B_1=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_1=-\frac{{\mathrm{μ}}_{0}i}{\sqrt{2}a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

Due to part 2 magnetic field is $B_2$

$$\begin{gathered} B_2=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}·a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_2=-\frac{3{\mathrm{μ}}_{0}i}{\sqrt{10}a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

Due to part 3 magnetic field is $B_3$

$$\begin{gathered} B_3=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}·3a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_3=-\frac{{\mathrm{μ}}_{0}i}{3\sqrt{10}a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

Due to part 4 magnetic field is $B_4$

$$\begin{gathered} B_4=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}·3a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{3a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_4=-\frac{\sqrt{2}{\mathrm{μ}}_{0}i}{6a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

Due to part 5 magnetic field is $B_5$

$$\begin{gathered} B_5=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}·3a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{3a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_5=-\frac{\sqrt{2}{\mathrm{μ}}_{0}i}{6a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

Due to part 6 magnetic field is $B_6$

$$\begin{gathered} B_6=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}·3a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_6=-\frac{{\mathrm{μ}}_{0}i}{3\sqrt{10}a\mathrm{π}}\stackrel{→}{\mathrm{k}} \end{gathered}$$

The net magnetic field at point Pthen becomes

$B=B_1+B_2+B_3+B_4+B_5+B_6+B_7+B_8$

$\begin{array}{rcl}\stackrel{→}{B}& =& -\stackrel{→}{k}\frac{{\mathrm{μ}}_{0}i}{a\mathrm{π}}·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{6}+\frac{1}{3\sqrt{10}}+\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{2}}\right)\\ \stackrel{→}{B}& =& -\stackrel{→}{k}\frac{2{\mathrm{μ}}_{0}i}{a\mathrm{π}}·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\\ \stackrel{→}{B}& =& -\stackrel{→}{k}\frac{8{\mathrm{μ}}_{0}i}{a4\mathrm{π}}·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\stackrel{→}{B}& =& \frac{-\stackrel{→}{k}\left(4\mathrm{π}×{10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A} ×10 \text{A}\right)}{4\mathrm{π}×10×{10}^{-2} \text{m}}\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\\ \stackrel{→}{B}& =& -\stackrel{→}{k}·{10}^{-4}\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right) \text{T}\\ \stackrel{→}{B}& =& -\left(2.0×{10}^{-4} \text{T}\right)\stackrel{→}{k}\end{array}$

Hence the net magnetic field at pointP is$\stackrel{→}{B}=-\left(2.0×{10}^{-4} \text{T}\right)\stackrel{→}{k}$