Chapter 30, Problem 28PQ

(a)

To determine

The magnitude and direction of magnetic field at point $A$.

Answer to Problem 28PQ

The magnetic field at the point $A$ is the magnetic field at the point $A$ is $-1.00\times {10}^{-5}\widehat{i}\text{T}$ .

Explanation of Solution

Refer to figure P30.28; the direction of the magnetic field can be described by the Right hand rule.

The Right Hand Thumb rule states that when a conductor is held such that the thumb points in the direction of the current flowing through the conductor, then the direction of curling of the fingers around the conductor shows the direction of magnetic field.

Write the expression for magnetic field at point $A$ due to the wire carrying current $I_1$  as.

  $\overrightarrow{B_1}=\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(-\widehat{i}\right)$                                                                                           (I)

Here, $d$ is the distance of the point $A$ from the wire carrying current $I_1$ , $I_1$ is the current flowing through the conductor and the field at point $A$ due to the wire carrying current $I_1$  is directed towards negative X-axis.

Write the expression for magnetic field at point $A$ due to the wire carrying current $I_2$  as.

  $\overrightarrow{B_2}=\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(-\widehat{j}\right)$                                                                                         (II)

Here, $d$ is the distance of the point $A$ from the wire carrying current $I_2$ , $I_2$ is the current flowing through the conductor and the field at point $A$ due to the wire carrying current $I_2$ is directed towards negative Y-axis.

Write the expression for magnetic field at point $A$ due to the wire carrying current $I_3$  as.

  $\overrightarrow{B_3}=\left(\frac{{\mu }_0{I}_3}{2\pi d}\right)\left(\widehat{j}\right)$                                                                                         (III)

Here, $d$ is the distance of the point $A$ from the wire carrying current $I_3$ , $I_3$ is the current flowing through the conductor and the field at point $A$ due to the wire carrying current $I_3$  is directed towards positive Y-axis.

Write the expression for the net magnetic field at point $A$ as.

  ${\overrightarrow{B}}_{net}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$                                                                                      (IV)

Here, ${\overrightarrow{B}}_{net}$ is the net magnetic field.

Conclusion:

Substitute $\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(-\widehat{i}\right)$ for $\overrightarrow{B_1}$, $\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(-\widehat{j}\right)$ for $\overrightarrow{B_2}$, $\left(\frac{{\mu }_0{I}_3}{2\pi d}\right)\left(\widehat{j}\right)$ for $\overrightarrow{B_3}$ in equation (IV).

    ${\overrightarrow{B}}_{net}=\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(-\widehat{i}\right)+\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(-\widehat{j}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi d}\right)\left(\widehat{j}\right)$

Substitute $4\pi \times {10}^{-7}\text{Tm/A}$ for ${\mu }_0$, $0.1\text{m}$ for $d$ and $5.00\text{A}$ for $I_1$ , $5\text{A}$ for $I_2$ and $5\text{A}$ for $I_3$ in above equation.

  $\begin{array}{c}{\overrightarrow{B}}_{net}=\left[\begin{array}{l}\left(\frac{\left(4\pi \times {10}^{-7}\text{}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(0.1\text{m}\right)}\right)\left(-\widehat{i}\right)+\left(\frac{\left(4\pi \times {10}^{-7}\text{}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(0.1\text{m}\right)}\right)\left(-\widehat{j}\right)\\ +\left(\frac{\left(4\pi \times {10}^{-7}\text{}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(0.1\text{m}\right)}\right)\left(\widehat{j}\right)\end{array}\right]\\ =\left(100\times {10}^{-7}\text{T}\right)\left(-\widehat{i}\right)\\ =-1.00\times {10}^{-5}\widehat{i}\text{T}\end{array}$

Thus, the magnetic field at the point $A$ is $-1.00\times {10}^{-5}\widehat{i}\text{T}$ .

(b)

To determine

The magnitude and direction of magnetic field at point $B$.

Answer to Problem 28PQ

The magnetic field at the point $B$ is $\underset{\_}{-1.50\times {10}^{-5}\widehat{i}\text{T}}$ .

Explanation of Solution

Refer to figure P30.28;

Write the expression for magnetic field at point $B$ due to the wire carrying current $I_1$  as.

  $\overrightarrow{B_1}=\left(\frac{{\mu }_0{I}_1}{2\pi \left(2d\right)}\right)\left(-\widehat{i}\right)$                                                                                     (V)

Here, $2d$ is the distance of the point $B$ from the wire carrying current $I_1$ and the field at point $B$ due to the wire carrying current $I_1$  is directed towards negative X-axis.

Consider the distance of point $B$ from the conductor carrying the current $I_2$ is $a$.

  $\begin{array}{c}a=\sqrt{d^2+d^2}\\ =\sqrt{2}d\end{array}$

Write the expression for magnetic field at point $B$ due to the wire carrying current $I_2$  as.

  $\overrightarrow{B_2}=\left(\left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\right)\cos 45°\right)\left(-\widehat{i}\right)+\left(\left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\right)\sin 45°\right)\left(-\widehat{j}\right)$                (VI)

Here, $\sqrt{2}d$ is the distance of the point $B$ from the wire carrying current $I_2$ and the field at point $B$ due to the wire carrying current $I_2$  is directed towards ${45}^{\circ }$ from negative X-axis to negative Y-axis.

Consider the distance of point $B$ from the conductor carrying the current $I_3$ is $b$.

  $\begin{array}{c}b=\sqrt{d^2+d^2}\\ =\sqrt{2}d\end{array}$

Write the expression for magnetic field at point $B$ due to the wire carrying current $I_3$  as.

  $\overrightarrow{B_3}=\left(\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\right)\sin 45°\right)\left(-\widehat{i}\right)+\left(\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\right)\cos 45°\right)\left(\widehat{j}\right)$                  (VII)

Here, $\sqrt{2}d$ is the distance of the point $B$ from the wire carrying current $I_3$ and the field at point $B$ due to the wire carrying current $I_3$  is directed towards ${45}^{\circ }$ from negative X-axis to positive Y-axis.

Write the expression for the net magnetic field at point $B$ as.

  ${\overrightarrow{B}}_{net}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$                                                                                    (VIII)

Here, ${\overrightarrow{B}}_{net}$ is the net magnetic field.

Conclusion:

Substitute $\left(\frac{{\mu }_0{I}_1}{2\pi \left(2d\right)}\right)\left(-\widehat{i}\right)$ for $\overrightarrow{B_1}$, $\left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\cos 45°\right)\left(-\widehat{i}\right)+\left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\sin 45°\right)\left(-\widehat{j}\right)$ for $\overrightarrow{B_2}$, $\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\sin 45°\right)\left(-\widehat{i}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\cos 45°\right)\left(\widehat{j}\right)$ for $\overrightarrow{B_3}$ in equation (VIII).

  ${\overrightarrow{B}}_{net}=\left[\begin{array}{l}\left(\frac{{\mu }_0{I}_1}{2\pi \left(2d\right)}\right)\left(-\widehat{i}\right)+\left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\cos 45°\right)\left(-\widehat{i}\right)+\\ \left(\frac{{\mu }_0{I}_2}{2\pi \left(\sqrt{2}d\right)}\sin 45°\right)\left(-\widehat{j}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\sin 45°\right)\left(-\widehat{i}\right)\\ +\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{2}d\right)}\cos 45°\right)\left(\widehat{j}\right)\end{array}\right]$

Substitute $4\pi \times {10}^{-7}\text{Tm/A}$ for ${\mu }_0$, $0.1\text{m}$ for $d$  and $5.00\text{A}$ for $I_1$ , $5.00\text{A}$ for $I_2$ and $5.00\text{A}$ for $I_3$ in above equation

  $\begin{array}{c}{\overrightarrow{B}}_{net}=\left[\begin{array}{l}\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)5.00\text{A}}{2\pi \left(0.2\text{ m}\right)}\right)\left(-\widehat{i}\right)+\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(\sqrt{2}\times 0.1\right)\text{m}}\cos 45°\right)\left(-\widehat{i}\right)\\ \left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(\sqrt{2}\left(0.1\right)\right)\text{m}}\sin 45°\right)\left(-\widehat{j}\right)+\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(\sqrt{2}\left(0.1\right)\right)\text{m}}\sin 45°\right)\left(-\widehat{i}\right)\\ +\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(\sqrt{2}\left(0.1\right)\right)\text{m}}\sin 45°\right)\left(\widehat{j}\right)\end{array}\right]\\ =\left(\frac{3\left(4\pi \times {10}^{-7}\text{Tm/A}\right)5.00\text{A}}{2\pi \left(0.2\text{ m}\right)}\right)\left(-\widehat{i}\right)\\ =\left(150\times {10}^{-7}\text{T}\right)\left(-\widehat{i}\right)\\ =-1.50\times {10}^{-5}\widehat{i}\text{T}\end{array}$

Thus, the magnetic field at the point $B$ is $\underset{\_}{-1.50\times {10}^{-5}\widehat{i}\text{T}}$ .

(c)

To determine

The magnitude and direction of magnetic field at point $C$.

Answer to Problem 28PQ

The magnetic field at the point $C$ is $\underset{\_}{\left[\left(1.2\times {10}^{-5}\right)\left(i\right)+\left(1.37\times {10}^{-5}\right)\left(\widehat{j}\right)\right]\text{T}}$.

Explanation of Solution

Refer to figure P30.28; write the expression for magnetic field at point $C$ due to the wire carrying current $I_1$  as.

  $\overrightarrow{B_1}=\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(\widehat{j}\right)$                                                                                          (IX)

Here, $d$ is the distance of the point $C$ from the wire carrying current $I_1$ and the field at point $C$ due to the wire carrying current $I_1$  is directed towards positive Y-axis.

Write the expression for magnetic field at point $A$ due to the wire carrying current $I_2$  as.

  $\overrightarrow{B_2}=\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(\widehat{i}\right)$                                                                                            (X)

Here, $d$ is the distance of the point $C$ from the wire carrying current $I_2$ , $I_2$ is the current flowing through the conductor and the field at point $C$ due to the wire carrying current $I_2$  is directed towards positive X-axis.

Consider the distance of point $B$ from the conductor carrying the current $I_3$ is $a$.

  $\begin{array}{c}a=\sqrt{d^2+{\left(2d\right)}^2}\\ =\sqrt{5}d\end{array}$

The angle made by the magnetic field vector from the X-axis is.

  $\begin{array}{c}\tan \theta =\frac{2d}{d}\\ \theta ={\tan }^{-1}2\\ =63.43°\end{array}$

Write the expression for magnetic field at point $B$ due to the wire carrying current $I_2$  as.

  $\overrightarrow{B_3}=\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\cos 63.43°\right)\left(\widehat{i}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\sin 63.43°\right)\left(\widehat{j}\right)$                   (XI)

Here, $\sqrt{5}d$ is the distance of the point $C$ from the wire carrying current $I_3$ and the field at point $C$ due to the wire carrying current $I_3$  is directed towards $63.43°$ from positive X-axis to positive Y-axis.

The net magnetic field at point $C$ is given as.

  ${\overrightarrow{B}}_{net}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$                                                                                     (XII)

Here, ${\overrightarrow{B}}_{net}$ is the net magnetic field.

Conclusion:

Substitute $\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(\widehat{j}\right)$ for $\overrightarrow{B_1}$, $\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(\widehat{i}\right)$ for $\overrightarrow{B_2}$ and $\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\cos 63.43°\right)\left(\widehat{i}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\sin 63.43°\right)\left(\widehat{j}\right)$ for $\overrightarrow{B_3}$ in equation (XII).

  ${\overrightarrow{B}}_{net}=\left[\begin{array}{l}\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)\left(\widehat{j}\right)+\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)\left(\widehat{i}\right)\\ +\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\cos 63.43°\right)\left(\widehat{i}\right)+\left(\frac{{\mu }_0{I}_3}{2\pi \left(\sqrt{5}d\right)}\sin 63.43°\right)\left(\widehat{j}\right)\end{array}\right]$

Substitute $4\pi \times {10}^{-7}\text{Tm/A}$ for ${\mu }_0$, $0.1\text{ m}$ for $d$, $5.00\text{ A}$ for $I_1$, $5.00\text{ A}$ for $I_2$ and $5.00\text{ A}$ for $I_3$ in above equation.

  $\begin{array}{c}{\overrightarrow{B}}_{net}=\left[\begin{array}{l}\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(0.1\text{ m}\right)}\right)\left(\widehat{j}\right)+\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)}{2\pi \left(0.1\text{ m}\right)}\right)\left(\widehat{i}\right)\\ +\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)\left(0.45\right)}{2\pi \left(\sqrt{5}\left(0.1\right)\right)\text{ m}}\right)\left(\widehat{i}\right)\\ +\left(\frac{\left(4\pi \times {10}^{-7}\text{Tm/A}\right)\left(5.00\text{A}\right)\left(0.84\right)}{2\pi \left(\sqrt{5}\left(0.1\right)\right)\text{ m}}\right)\left(\widehat{j}\right)\end{array}\right]\\ =\left[\left({10}^{-5}\right)\left(\widehat{j}\right)+\left({10}^{-5}\right)\left(\widehat{i}\right)+\left(0.2\times {10}^{-5}\right)\left(\widehat{i}\right)+\left(0.37\times {10}^{-5}\right)\left(\widehat{j}\right)\right]\text{T}\\ =\left[\left(1.2\times {10}^{-5}\right)\left(\widehat{i}\right)+\left(1.37\times {10}^{-5}\right)\left(\widehat{j}\right)\right]\text{T}\end{array}$

Thus, the magnetic field at the point $C$ is $\underset{\_}{\left[\left(1.2\times {10}^{-5}\right)\left(i\right)+\left(1.37\times {10}^{-5}\right)\left(\widehat{j}\right)\right]\text{T}}$ .