Question

Equation 29-4 gives the magnitude Bof the magnetic field set up by a current in an infinitely long straight wire, at a point Pat perpendicular distance R from the wire. Suppose that point P is actually at perpendicular distance Rfrom the midpoint of a wire with a finite length L.Using Eq. 29-4 to calculate Bthen results in a certain percentage error. What value must the ratio $\frac{\mathbf{L}}{\mathbf{R}}$exceed if the percentage error is to be less than $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}\mathbf{?}$ That is, what $\frac{\mathbf{L}}{\mathbf{R}}$gives

$\frac{\left(\mathbf{B}\mathbf{}\mathbf{from}\mathbf{}\mathbf{Eq}\mathbf{.}\mathbf{}\mathbf{29}\mathbf{-}\mathbf{4}\right)\mathbf{-}\left(\mathbf{B}\mathbf{}\mathbf{actual}\right)}{\left(\mathbf{B}\mathbf{}\mathbf{actual}\right)}\mathbf{}\left(\mathbf{100}\mathbf{\%}\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}\mathbf{?}$

Short answer

The exceeded value of the ratio $\frac{L}{R}$if the percentage error is to be less than$1.00\%$ is $14.1$.

Step-by-step solution

Given

$\frac{B_{\infty }-B}{B}=1.00\%$

Understanding the concept

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:

$$\begin{gathered} B_{\infty }=\frac{{\mu }_{0}i}{2\pi R} \\ dB=\frac{{\mu }_{0}idl}{4\pi }\frac{\sin \theta }{r^2} \end{gathered}$$

Calculate the exceeded value of the ratio LR if the percentage error is to be less than 1.00 %

According to Bio-Savarts law

$dB=\frac{{\mu }_{0}idl}{4\pi }\frac{\sin \theta }{r^2}$

If r makes an angle$\theta$ with l, then

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

And

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

Integrating an equation of Bio-Savarts law

$$\begin{gathered} B=\int dB \\ =\int \frac{{\mu }_{0}idL}{4\pi }\frac{\sin \theta }{r^2} \\ B=\frac{{\mu }_{0}i}{4\pi }\int \frac{R}{\sqrt{{\left(\frac{L}{2}\right)}^2+R^2}}\frac{dL}{{\left(\sqrt{{\left(\frac{L}{2}\right)}^2+R^2}\right)}^2} \\ B=\frac{{\mu }_{0}i}{4\pi }\int \frac{\sin \theta dL}{r^2} \\ B=\frac{{\mu }_{0}iR}{4\pi }\int \frac{dl}{{\left({\left(\frac{L}{2}\right)}^2+R^2\right)}^{3/2}} \end{gathered}$$

Solve further as:

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

Since:

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

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$14.1$ .