91Ó°ÊÓ

$\frac{B_{∞}-B}{B}=1.00\%$

Use the formula of Biot-Savarts law and the formula of magnetic field in the given ratio of magnetic field to find the exceeded value of the ratio $\frac{\mathit{L}}{\mathit{R}}$if the percentage error is to be less than$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\%}$

Formula:

$$\begin{gathered} B_{∞}=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R} \\ dB=\frac{{\mathrm{μ}}_{0}idl}{4\mathrm{π}}\frac{\sin \mathrm{θ}}{r^2} \end{gathered}$$

According to Bio-Savarts law

$dB=\frac{{\mathrm{μ}}_{0}idl}{4\mathrm{π}}\frac{\sin \mathrm{θ}}{r^2}$

If r makes an angle$\mathrm{θ}$ with l, then

$r=\sqrt{l^2+R^2}$

And

$\sin \mathrm{θ}=\frac{R}{R}=\frac{R}{\sqrt{{\left(\frac{L}{2}\right)}^2+R^2}}$

Integrating an equation of Bio-Savarts law

$$\begin{gathered} B=∫dB \\ =∫\frac{{\mathrm{μ}}_{0}idL}{4\mathrm{π}}\frac{\sin \mathrm{θ}}{r^2} \\ B=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}}∫\frac{R}{\sqrt{{\left(\frac{L}{2}\right)}^2+R^2}}\frac{dL}{{\left(\sqrt{{\left(\frac{L}{2}\right)}^2+R^2}\right)}^2} \\ B=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}}∫\frac{\sin \mathrm{θ}dL}{r^2} \\ B=\frac{{\mathrm{μ}}_{0}iR}{4\mathrm{π}}∫\frac{dl}{{\left({\left(\frac{L}{2}\right)}^2+R^2\right)}^{3/2}} \end{gathered}$$

Solve further as:

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}iR}{4\mathrm{π}}\frac{1}{R^2}\left[\frac{l}{{\left({\left(\frac{L}{2}\right)}^2+R^2\right)}^{\frac{1}{2}}}\right] \\ B=\frac{{\displaystyle {\mathrm{μ}}_{0}i}}{{\displaystyle 2\mathrm{π}R}}\left[\frac{{\displaystyle l}}{{\displaystyle {\left({\left(\frac{{\displaystyle L}}{{\displaystyle 2}}\right)}^2+R^2\right)}^{1/2}}}\right] \end{gathered}$$

Since:

$$\begin{gathered} \frac{B_{∞}-B}{B}=1\% \\ \frac{B_{∞}-B}{B}=0.01 \\ \frac{B_{∞}}{B}-1=0.01 \\ \frac{B_{∞}}{B}=1.01 \\ \frac{\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}}{\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}\left[\frac{\frac{L}{2}}{{\left({\left(\frac{L}{2}\right)}^2+R^2\right)}^{\frac{1}{2}}}\right]}=1.01 \end{gathered}$$

Solve further as:

$$\begin{gathered} 1+\frac{4R^2}{L^2}=1.022 \\ \frac{4R^2}{L^2}=0.022 \\ \frac{L}{R}=13.5 \\ =14 \end{gathered}$$

Therefore, the exceeded value of the ratio L/R if the percentage error is to be less than$1.00\%$ is$14.1$ .