91Ó°ÊÓ

• The length of the rectangular loop is L.
• The width of therectangularloop isW.
• Current through the rectangular loop is i.

For this problem, we will use the equation for the magnetic field due to a finite wire of length L, carrying current i, at a point and distance R from the wire. The direction of the magnetic field will depend on the direction of the current. We can determine the direction by the right-hand thumb rule.

Formula:

$B_p=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}R}·\left(\frac{l}{{\left(L^2+R^2\right)}^{\frac{1}{2}}}\right)$

We break the rectangular wire loop into eight parts, as shown in the figure. By numbering them, we will find the magnetic field due to each part at the center of the loop.

Due to part 1, the magnetic field is,

$\begin{array}{rcl}{B}_1& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}L}{2}}·\frac{\left(\frac{W}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\\ & & \end{array}$

Due to part 2, the magnetic field is,

$\begin{array}{rcl}{B}_2& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}L}{2}}·\frac{\left(\frac{W}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\\ & & \end{array}$

.

Due to part 3, the magnetic field is,

$$\begin{gathered} B_3=\frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}W}{2}}·\frac{\left(\frac{L}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}} \\ =\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}} \end{gathered}$$

Due to part 4, the magnetic field is,

$\begin{array}{rcl}{B}_4& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}W}{2}}·\frac{\left(\frac{L}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\\ & & \end{array}$

Due to part 5, the magnetic field is,

$\begin{array}{rcl}{B}_5& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}L}{2}}·\frac{\left(\frac{W}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\end{array}$

Due to part 6, the magnetic field is,

$\begin{array}{rcl}{B}_6& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}L}{2}}·\frac{\left(\frac{W}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\\ & & \end{array}$

Due to part 7, the magnetic field is,

$\begin{array}{rcl}{B}_7& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}W}{2}}·\frac{\left(\frac{L}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\end{array}$

Due to part 8, the magnetic field is,

$\begin{array}{rcl}{B}_8& =& \frac{{\mathrm{μ}}_{0}i}{\frac{4\mathrm{π}W}{2}}·\frac{\left(\frac{L}{2}\right)}{{\left({\frac{W}{4}}^2+{\frac{L}{4}}^2\right)}^{\frac{1}{2}}}\\ & =& \frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\end{array}$

The net magnetic field at point C then becomes,

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

$$\begin{gathered} B=\frac{2{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{2{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{2{\mathrm{μ}}_{0}i}{2\mathrm{π}L}·\frac{W}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{2{\mathrm{μ}}_{0}i}{2\mathrm{π}W}·\frac{L}{{\left(W^2+L^2\right)}^{\frac{1}{2}}} \\ B=\frac{{\mathrm{μ}}_{0}i}{\mathrm{π}}·\left(\frac{W}{L{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{W}{L{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{L}{W{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{L}{W{\left(W^2+L^2\right)}^{\frac{1}{2}}}\right) \\ B=\frac{{\mathrm{μ}}_{0}i}{\mathrm{π}}·\left(\frac{2W}{L{\left(W^2+L^2\right)}^{\frac{1}{2}}}+\frac{2L}{W{\left(W^2+L^2\right)}^{\frac{1}{2}}}\right) \\ B=\frac{2{\mathrm{μ}}_{0}i}{\mathrm{π}}·\frac{1}{{\left(W^2+L^2\right)}^{\frac{1}{2}}}\left(\frac{W}{L}+\frac{L}{W}\right) \\ B=\frac{2{\mathrm{μ}}_{0}i}{\mathrm{π}}·\frac{{\left(L^2+W^2\right)}^{\frac{1}{2}}}{LW} \end{gathered}$$

Hence, it is proved that.