Chapter 19, Problem 67AP

a)

To determine

The direction and magnitude of the magnetic field at point $P$.

Answer to Problem 67AP

The net magnetic field at the point $P$ is $5.00\times {10}^{-7}\text{T}$and is directed out of the page.

Explanation of Solution

Given info:

The wire along the x-axis carries current of $5.00\text{A}$. The current along the y-axis carries current of $3.00\text{A}$. The distance of the point $P$ from x-axis is $40.0\text{cm}$, where as from y-axis is $30.0\text{cm}$.

Explanation:

The arrangement of the wires is shown in the following figure.

The magnetic field of a long straight wire is given by,

$B=\frac{{\mu }_{0}I}{2\pi r}$

• ${\mu }_0$ is the permeability of free space
• $I$ is the current
• $r$ is the distance of the point from the straight conductor

Consider the positive field to be out of the plane.

The magnetic field at point $P$ due to the wire along x-axis will be out of the page, whereas due to the wire along y-axis it will into the page.

The magnetic field at $P$ due to the wire along x-axis will be,

$B_x=\frac{{\mu }_0{I}_x}{2\pi r_x}$

• $I_x$ is the current in the wire along the x-axis
• $r_x$ is the distance of the point $P$ from the x-axis

$B_x$ is positive since it is pointing out of the page.

The magnetic field at $P$ due to the wire along y-axis will be,

$B_y=-\frac{{\mu }_0{I}_y}{2\pi r_y}$

• $I_y$ is the current in the wire along the y-axis
• $r_y$ is the distance of the point $P$ from the y-axis

$B_y$ is negative since it is pointing into the page.

The net magnetic field at the point $P$ will be,

$\begin{array}{c}{B}_{net}=\frac{{\mu }_0{I}_x}{2\pi r_x}-\frac{{\mu }_0{I}_y}{2\pi r_y}\\ =\frac{{\mu }_0}{2\pi }\left(\frac{I_x}{r_x}-\frac{I_y}{r_y}\right)\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $5.00\text{A}$ for $I_x$, $3.00\text{A}$ for $I_y$, $40.0\text{cm}$ for $r_x$, $30.0\text{cm}$ for $r_y$, to find the net magnetic field at $P$,

$\begin{array}{c}{B}_{net}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)}{2\pi }\left(\frac{5.00\text{A}}{40.0\text{cm}}-\frac{3.00\text{A}}{30.0\text{cm}}\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)}{2\pi }\left(\frac{5.00\text{A}}{40.0\times {10}^{-2}\text{m}}-\frac{3.00\text{A}}{30.0\times {10}^{-2}\text{m}}\right)\\ =5.00\times {10}^{-7}\text{T}\\ \text{=0}\text{.500}\text{μΤ}\end{array}$

Conclusion: The net magnetic field at the point $P$ is $5.00\times {10}^{-7}\text{T}$and is directed out of the page.

b)

To determine

The magnetic field at a point $30.0\text{cm}$ above the point of intersection of the wires.

Answer to Problem 67AP

At a point $30.0\text{cm}$ above the point of intersection of the wires the magnitude of the magnetic field is $\text{3}\text{.88}\text{μΤ}$ and this makes $-59.0°$ with the x-axis.

Explanation of Solution

Given info:

The wire along the x-axis carries current of $5.00\text{A}$. The current along the y-axis carries current of $3.00\text{A}$. The distance of the point $P$ from x-axis is $40.0\text{cm}$, where as from y-axis is $30.0\text{cm}$.

Explanation:

The system is shown in the figure 1 below.

At a point $30.0\text{cm}$ above the point of intersection of the wires the x-component of the magnetic field is due to the wire along y-axis, the y-component of the magnetic field is due to the wire along x-axis.

Consider the positive direction of the magnetic field to be out of the page.

The x-component of the magnetic field is,

$B_x=\frac{{\mu }_0{I}_y}{2\pi r}$

• $I_y$ is the current in the wire along the y-axis
• $r$ is the distance of the point from the intersection of the two wires

The y-component of the magnetic field is,

$B_y=-\frac{{\mu }_0{I}_x}{2\pi r}$

• $I_x$ is the current in the wire along the x-axis
• $r$ is the distance of the point from the intersection of the two wires

The magnitude of the net magnet field is,

$\begin{array}{c}{B}_{net}=\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2}\\ =\sqrt{{\left(\frac{{\mu }_0{I}_y}{2\pi r}\right)}^2+{\left(-\frac{{\mu }_0{I}_x}{2\pi r}\right)}^2}\\ =\left(\frac{{\mu }_0}{2\pi r}\right)\sqrt{{\left(I_y\right)}^2+{\left(I_x\right)}^2}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $5.00\text{A}$ for $I_x$, $3.00\text{A}$ for $I_y$, $30.0\text{cm}$ for $r$, to find the net magnetic field,

$\begin{array}{c}{B}_{net}=\left(\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}}{2\pi \left(30.0\text{cm}\right)}\right)\sqrt{{\left(3.00\text{A}\right)}^2+{\left(5.00\text{A}\right)}^2}\\ =3.88\times {10}^{-6}\text{T}\\ \text{=3}\text{.88}\text{μΤ}\end{array}$

The direction between the resultant magnetic field and the x-axis is,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{B_y}{B_x}\right)\\ ={\tan }^{-1}\left(\frac{-\frac{{\mu }_0{I}_x}{2\pi r}}{\frac{{\mu }_0{I}_y}{2\pi r}}\right)\\ ={\tan }^{-1}\left(-\frac{I_x}{I_y}\right)\end{array}$

Substitute $5.00\text{A}$ for $I_x$, $3.00\text{A}$ for $I_y$ to determine the angle,

$\begin{array}{c}\theta ={\tan }^{-1}\left(-\frac{5.00\text{A}}{3.00\text{A}}\right)\\ =-59.0°\end{array}$

Conclusion: At a point $30.0\text{cm}$ above the point of intersection of the wires the magnitude of the magnetic field is $\text{3}\text{.88}\text{μΤ}$ and this makes $-59.0°$ with the x-axis.