Chapter 30, Problem 20P

(a)

To determine

The magnitude and the direction of the net magnetic field at the mid way between the wires.

Answer to Problem 20P

The magnitude is $\text{4}\text{.0}\times {\text{10}}^{-6}\text{T}$ and the direction of the net magnetic field is into the page as $B_2>B_1$ at the mid way between the wires.

Explanation of Solution

Write the expression to obtain the magnetic field along the conductor.

    $B=\frac{{\mu }_{0}i}{2\pi r}$

Here, $B$ is the magnetic field, ${\mu }_0$ is the magnetic permittivity, $i$ is the current and $r$ is the distance between the conductor and the point where the magnetic field is to be determined.

The net magnetic field at the mid way between the wires is as shown in the figure below.

Figure-(1)

Write the expression to obtain the magnetic field due to the wire $1$ at the midway between the wires.

    $B_1=\frac{{\mu }_0{i}_1}{2\pi r_1}$

Here, $B_1$ is the magnetic field due to the wire $1$ at the midway between the wires and $r_1$ is the distance between the wire $1$ and the point where magnetic field is to be determined.

Substitute $\frac{d}{2}$ for $r_1$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{{\mu }_0{i}_1}{2\pi \left(\frac{d}{2}\right)}\\ =\frac{{\mu }_0{i}_1}{\pi d}\end{array}$                                                                                                              (I)

Write the expression to obtain the magnetic field due to the wire $2$ at the midway between the wires.

    $B_2=\frac{{\mu }_0{i}_2}{2\pi r_2}$

Here, $B_2$ is the magnetic field due to the wire $2$ at the midway between the wires and $r_2$ is the distance between the wire $2$ and the point where magnetic field is to be determined.

Substitute $\frac{d}{2}$ for $r_2$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{{\mu }_0{i}_2}{2\pi \left(\frac{d}{2}\right)}\\ =\frac{{\mu }_0{i}_2}{\pi d}\end{array}$                                                                                                            (II)

Write the expression to obtain the net magnetic field at the mid way between the wires.

    $B_{\text{net}}=B_2-B_{{}_1}$                                                                                                          (III)

Here, $B_{\text{net}}$ is the net magnetic field at the mid way between the wires.

Conclusion:

Substitute $3.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $20.0\text{cm}$ for $d$ in equation (I) to calculate $B_1$.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(3.00\text{A}\right)}{\pi \left(20.0\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(3.00\text{A}\right)}{\pi \left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =6.0\times {10}^{-6}\text{T}\end{array}$

Substitute $5.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $20.0\text{cm}$ for $d$ in equation (II) to calculate $B_2$.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{\pi \left(20.0\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{\pi \left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.0\times {10}^{-5}\text{T}\end{array}$

Substitute $6.0\times {10}^{-6}\text{T}$ for $B_1$ and $1.0\times {10}^{-5}\text{T}$ for $B_2$ in equation (III) to calculate $B_{\text{net}}$.

    $\begin{array}{c}{B}_{\text{net}}=1.0\times {10}^{-5}\text{T}-6.0\times {10}^{-6}\text{T}\\ =\text{4}\text{.0}\times {\text{10}}^{-6}\text{T}\end{array}$

Therefore, the magnitude is $\text{4}\text{.0}\times {\text{10}}^{-6}\text{T}$ and the direction of the net magnetic field is into the page as $B_2>B_1$ at the mid way between the wires.

(b)

To determine

The magnitude and the direction of the net magnetic field at point $P$ which is located at a distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current.

Answer to Problem 20P

The magnitude of net magnetic field at point $P$ is $6.68\times {\text{10}}^{-6}\text{T}$ and its direction is $13°$ to the negative x-axis.

Explanation of Solution

The net magnetic field at point $P$ which is located at a distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current is as shown below.

Figure-(2)

Write the expression to obtain the magnetic field due to the wire $1$ at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current.

    $B_1=\frac{{\mu }_0{i}_1}{2\pi r_1}$

Here, $B_1$ is the magnetic field due to the wire $1$ at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current and $r_1$ is the distance between the wire $1$ and the point where magnetic field is to be determined.

Substitute $\sqrt{d^2+d^2}$ for $r_1$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{{\mu }_0{i}_1}{2\pi \sqrt{d^2+d^2}}\\ =\frac{{\mu }_0{i}_1}{\pi \sqrt{2d^2}}\\ =\frac{{\mu }_0{i}_1}{\pi d\sqrt{2}}\end{array}$                                                                                                   (IV)

Write the expression to obtain the magnetic field due to the wire $2$ at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current.

    $B_2=\frac{{\mu }_0{i}_2}{2\pi r_2}$

Here, $B_2$ is the magnetic field due to the wire $2$ at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current and $r_2$ is the distance between the wire $2$ and the point where magnetic field is to be determined.

Substitute $d$ for $r_2$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{{\mu }_0{i}_2}{2\pi d}\\ =\frac{{\mu }_0{i}_2}{2\pi d}\end{array}$                                                                                                                 (V)

Write the expression to obtain the net magnetic field at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current.

    ${\overrightarrow{B}}_{\text{net}}={\overrightarrow{B}}_1+{\overrightarrow{B}}_{{}_2}$                                                                                                          (VI)

Here, $B_{\text{net}}$ is the net magnetic field at the mid way between the wires.

Write the expression to obtain the magnitude of the net magnetic field at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current.

    $\left|B_{\text{net}}\right|=\sqrt{B_{\text{x}}^2+B_{\text{y}}^2}$                                                                                                   (VII)

Here, $\left|B_{\text{net}}\right|$ is the magnitude of the net magnetic field at the distance $20.0\text{cm}$ above the wire carrying $5.00\text{A}$ current, $B_{\text{x}}$ is the x-component of the net magnetic field and $B_{\text{y}}$ is the y-component of the net magnetic field.

Write the expression to obtain direction of the net magnetic field at point $P$ with reference to x-axis.

    $\theta ={\tan }^{-1}\left(\frac{B_{\text{y}}}{B_{\text{x}}}\right)$                                                                                                    (VIII)

Here, $\theta$ is the direction of the net magnetic field at point $P$ with reference to x-axis.

Conclusion:

Substitute $3.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $20.0\text{cm}$ for $d$ in equation (IV) to calculate $B_1$.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(3.00\text{A}\right)}{\pi \left(20.0\text{cm}\right)\sqrt{2}}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(3.00\text{A}\right)}{\pi \left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{2}}\\ =2.14\times {10}^{-6}\text{T}\end{array}$

The value of $B_1$ in x and y component is,

    ${\overrightarrow{B}}_1=B_1\sin 45°\left(-\widehat{i}\right)+B_1\cos 45°\left(\widehat{j}\right)$

Substitute $2.14\times {10}^{-6}\text{T}$ for $B_1$ in the above equation.

    $\begin{array}{c}{\overrightarrow{B}}_1=\left(2.14\times {10}^{-6}\text{T}\right)\sin 45°\left(-\widehat{i}\right)+\left(2.14\times {10}^{-6}\text{T}\right)\cos 45°\left(\widehat{j}\right)\\ =-\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)+\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)\end{array}$

Substitute $5.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $20.0\text{cm}$ for $d$ in equation (V) to calculate $B_2$.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{2\pi \left(20.0\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{2\pi \left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =5.0\times {10}^{-6}\text{T}\end{array}$

The value of $B_2$ in x and y component is,

    ${\overrightarrow{B}}_2=B_2\left(-\widehat{i}\right)+\left(0\right)\left(\widehat{j}\right)$

Substitute $5.0\times {10}^{-6}\text{T}$ for $B_2$ in the above equation.

    ${\overrightarrow{B}}_2=-\left(5.0\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)$

Substitute $\left[-\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)+\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)\right]$ for ${\overrightarrow{B}}_1$ and $-\left(5.0\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)$ for ${\overrightarrow{B}}_2$ in equation (VI) to calculate ${\overrightarrow{B}}_{\text{net}}$.

    $\begin{array}{c}{\overrightarrow{B}}_{\text{net}}=-\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)+\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)-\left(5.0\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)\\ =-\left(6.51\times {10}^{-6}\text{T}\right)\left(\widehat{i}\right)+\left(1.51\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)\end{array}$

Substitute $-\left(6.51\times {10}^{-6}\text{T}\right)$ for $B_{\text{x}}$ and $\left(1.51\times {10}^{-6}\text{T}\right)$ for $B_{\text{y}}$ in equation (VII) to calculate $\left|B_{\text{net}}\right|$

    $\begin{array}{c}\left|B_{\text{net}}\right|=\sqrt{{\left(-\left(6.51\times {10}^{-6}\text{T}\right)\right)}^2+{\left(\left(1.51\times {10}^{-6}\text{T}\right)\right)}^2}\\ =6.68\times {10}^{-6}\text{T}\end{array}$

Substitute $\left(6.51\times {10}^{-6}\text{T}\right)$ for $B_{\text{x}}$ and $\left(1.51\times {10}^{-6}\text{T}\right)$ for $B_{\text{y}}$ in equation (VIII) to calculate $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(1.51\times {10}^{-6}\text{T}\right)}{\left(6.51\times {10}^{-6}\text{T}\right)}\right)\\ =13°\end{array}$

Therefore, the magnitude of net magnetic field at point $P$ is $6.68\times {\text{10}}^{-6}\text{T}$ and its direction is $13°$ to the negative x-axis.