Chapter 30, Problem 75PQ

To determine

The distance at which the approximate value equals the exact value.

Answer to Problem 75PQ

The approximate value starts approaching the exact value at $\underset{\_}{2R=0.05\text{m}}$ and matches completely around $\underset{\_}{3R=0.075\text{m}}$.

Explanation of Solution

Write the expression for the approximate value of force as.

  $\overrightarrow{B}\approx \frac{{\mu }_{0}I{R}^2}{2y^3}\widehat{j}$                                                                                                   (I)

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current flowing, $R$ is the radius of the loop and $y$ is the distance of the point from the loop.

Write the expression for the exact value of force as.

  $\overrightarrow{B}=\frac{{\mu }_{0}I{R}^2}{2{\left(R^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\widehat{j}$                                                                                     (II)

Conclusion:

Check for the different values of y varying from $2R$ to $4R$.

Substitute $\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)$ for ${\mu }_0$, $0.50\text{A}$ for $I$ , $2.5\text{cm}$ for $R$ and $2R$ for $y$ in equation (I).

  $\begin{array}{c}\overrightarrow{B}\approx \frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\right)}^2}{2{\left(2\left(2.5\text{cm}\right)\right)}^3}\widehat{j}\\ \approx \frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{2{\left(2\left(2.5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^3}\widehat{j}\\ \approx 1.57\times {10}^{-6}\text{T}\widehat{j}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}$ for ${\mu }_0$, $0.50\text{A}$ for $I$ , $2.5\text{cm}$ for $R$ and $2R$ for $y$ in equation (II).

  $\begin{array}{c}\overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\right)}^2}{2{\left\{{\left(2.5\text{cm}\right)}^2+\left(4{\left(2.5\text{cm}\right)}^2\right)\right\}}^3}\widehat{j}\\ =\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{2{\left\{{\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2+\left(4{\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\right\}}^3}\widehat{j}\\ =1.13\times {10}^{-6}\text{T}\widehat{j}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}$ for ${\mu }_0$, $0.50\text{A}$ for $I$ , $2.5\text{cm}$ for $R$ and $3R$ for $y$ in equation (I).

  $\begin{array}{c}\overrightarrow{B}\approx \frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\right)}^2}{2{\left(3\left(2.5\text{cm}\right)\right)}^3}\widehat{j}\\ \approx \frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{2{\left(3\left(2.5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^3}\widehat{j}\\ \approx 4.65\times {10}^{-7}\text{T}\widehat{j}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}$ for ${\mu }_0$, $0.50\text{A}$ for $I$ , $2.5\text{cm}$ for $R$ and $3R$ for $y$ in equation (II).

  $\begin{array}{c}\overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\right)}^2}{2{\left\{{\left(2.5\text{cm}\right)}^2+\left(9{\left(2.5\text{cm}\right)}^2\right)\right\}}^3}\widehat{j}\\ =\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}\right)\left(0.50\text{A}\right){\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{2{\left\{{\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2+\left(9{\left(2.5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\right\}}^3}\widehat{j}\\ =3.97\times {10}^{-7}\text{T}\widehat{j}\end{array}$

The values of approximate and exact magnetic fields are as tabulated below.

y (in m)	B (approximate) (in $10^{-5}T$)	B (exact) (in $10^{-5}T$)
$0.025$	$1.256$	$0.444$
$0.050$	$0.157$	$0.113$
$0.075$	$0.046$	$0.040$
$0.10$	$0.019$	$0.017$
0.125	$0.010$	$0.095$

Thus, the approximate value starts approaching the exact value at $\underset{\_}{2R=0.05\text{m}}$ and matches completely around $\underset{\_}{3R=0.075\text{m}}$.