Chapter 30, Problem 30.19P

Determine the magnetic field (in terms of I, a, and d) at the origin due to the current loop in Figure P29.9. The loop extends to infinity above the figure.

Figure P29.9

To determine

The magnetic field at the origin due to current loop.

Answer to Problem 30.19P

The magnetic field at the origin due to current loop is $\frac{{\mu }_{o}I}{2\pi ad}\left[\sqrt{d^2+a^2}-d\right]$.

Explanation of Solution

Given Info: current flowing through the loop is $I$, distance of point $O$ from loop is $d$ and length of horizontal part of current loop from origin is $a$.

Explanation:

Formula to calculate magnetic field given by Biot savarts law,

$dB=\frac{{\mu }_{o}I}{4\pi }\frac{dl\sin \theta }{r^2}\stackrel{\wedge }{r}$ (1)

Formula to calculate $r$ from Pythagorean theorem,

$r=\sqrt{a^2+d^2}$

Write the expression to calculate $r$ from the figure (1),

$\begin{array}{c}\text{cosec}\theta =\frac{r}{a}\\ r=a\text{cosec}\theta \end{array}$

Similarly,

$\begin{array}{c}\tan \theta =-\frac{a}{l}\\ l=-a\cot \theta \end{array}$

Differentiate above equation,

$\begin{array}{c}dl=d\left(-a\cot \theta \right)\\ =-a\left[d\left(\cot \theta \right)\right]\\ =-a\left(-{\text{cosec}}^2\theta \right)d\theta \\ =a{\text{cosec}}^2\theta d\theta \end{array}$

Substitute $a\cos e{c}^2\theta d\theta$ for $dl$ and $a\cos ec\theta$ for $r$ in equation (1)

$\begin{array}{c}dB=\frac{{\mu }_{o}I}{4\pi }\frac{\left(a{\text{cosec}}^2\theta d\theta \right)\sin \theta }{{\left(a\text{cosec}\theta \right)}^{}{}^2}\widehat{\text{r}}\\ dB=\frac{{\mu }_{o}I}{4\pi a}\sin \theta d\theta \widehat{\text{r}}\end{array}$ (2)

On integrate both sides from limit ${\theta }_{1}to{\theta }_2$.

$\begin{array}{c}{\displaystyle \int dB}={\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\frac{{\mu }_{o}I}{4\pi a}\sin \theta d\theta \stackrel{\wedge }{r}}\\ B=\frac{{\mu }_{o}I}{4\pi a}{\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\sin \theta d\theta \stackrel{\wedge }{r}}\\ =\frac{{\mu }_{o}I}{4\pi a}{\left[-\cos \right]}_{{\theta }_1}^{{\theta }_2}\\ =\frac{{\mu }_{o}I}{4\pi a}\left[\cos {\theta }_1-\cos {\theta }_2\right]\end{array}$ (3)

For length left to point $O$, ${\theta }_2=0$ and $\cos {\theta }_1=\frac{d}{\sqrt{d^2+a^2}}$.

$\begin{array}{c}{B}_{\text{OA}}=\frac{{\mu }_{o}I}{4\pi a}\left[\frac{d}{\sqrt{d^2+a^2}}-\cos 0\right]\\ =\frac{{\mu }_{o}I}{4\pi a}\left[\frac{d}{\sqrt{d^2+a^2}}-1\right]\end{array}$

For length between $O$ and loop, $\cos {\theta }_1=\frac{a}{\sqrt{d^2+a^2}}$ and $\cos {\theta }_2=-\frac{a}{\sqrt{d^2+a^2}}$.

$B_{\text{A}B}=\frac{{\mu }_{o}I}{4\pi d}\left[\frac{a}{\sqrt{d^2+a^2}}+\frac{a}{\sqrt{d^2+a^2}}\right]$

For length right to point $O$, $\cos {\theta }_1=-\frac{d}{\sqrt{d^2+a^2}}$ and $\cos {\theta }_2=180$.

$\begin{array}{c}{B}_{BC}=\frac{{\mu }_{o}I}{4\pi a}\left[\cos 180-\left(-\frac{d}{\sqrt{d^2+a^2}}\right)\right]\\ =\frac{{\mu }_{o}I}{4\pi a}\left[-1+\frac{d}{\sqrt{d^2+a^2}}\right]\end{array}$

Net magnetic field at point $O$ is,

$B_O=B_{\text{OA}}+B_{AB}+B_{BC}$

Substitute $\frac{{\mu }_{o}I}{4\pi a}\left[-1+\frac{d}{\sqrt{d^2+a^2}}\right]$ for $B_{\text{OA}}$, $\frac{{\mu }_{o}I}{4\pi d}\left[\frac{a}{\sqrt{d^2+a^2}}+\frac{a}{\sqrt{d^2+a^2}}\right]$ for $B_{AB}$ and $\frac{{\mu }_{o}I}{4\pi a}\left[\frac{d}{\sqrt{d^2+a^2}}-1\right]$ for $B_{BC}$ in above equation.

$\begin{array}{c}{B}_{\text{O}}=\frac{{\mu }_{o}I}{4\pi a}\left[-1+\frac{d}{\sqrt{d^2+a^2}}\right]+\frac{{\mu }_{o}I}{4\pi d}\left[\frac{a}{\sqrt{d^2+a^2}}+\frac{a}{\sqrt{d^2+a^2}}\right]+\frac{{\mu }_{o}I}{4\pi a}\left[+\frac{d}{\sqrt{d^2+a^2}}-1\right]\\ =\frac{{\mu }_{o}I}{4\pi }\left[-\frac{1}{a}-\frac{d}{a\sqrt{d^2+a^2}}\right]+\left[\frac{a}{d\sqrt{d^2+a^2}}+\frac{a}{d\sqrt{d^2+a^2}}\right]+\left[-\frac{d}{a\sqrt{d^2+a^2}}-\frac{1}{a}\right]\\ =\frac{{\mu }_{o}I}{4\pi }\left[\frac{-d\sqrt{d^2+a^2}+d^2+2a^2+d^2-d\sqrt{d^2+a^2}}{ad\sqrt{d^2+a^2}}\right]\\ =\frac{{\mu }_{o}I}{2\pi }\left[\frac{d^2+a^2-d\sqrt{d^2+a^2}}{ad\sqrt{d^2+a^2}}\right]\\ =\frac{{\mu }_{o}I}{2\pi ad}\left[\sqrt{d^2+a^2}-d\right]\end{array}$

Conclusion:

Therefore, magnetic field at point $O$ is $\frac{{\mu }_{o}I}{2\pi ad}\left[\sqrt{d^2+a^2}-d\right]$.