Chapter 30, Problem 30.72CP

(a)

To determine

To draw: The magnetic field pattern in the $yz$ plane.

Answer to Problem 30.72CP

The magnetic field pattern in the $yz$ plane is shown below.

Figure (1)

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

The magnetic field pattern in the $yz$ plane is shown below.

Figure (1)

From the right hand thumb rule, when the thumb is directed towards the direction of current, the curled fingers show the direction of magnetic field.

(b)

To determine

The value of magnetic field at origin.

Answer to Problem 30.72CP

The value of magnetic field at origin is $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$.

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

The direction of magnetic field due to current carry wire is shown below.

Figure (2)

Write the expression for the magnetic field due to current carrying wire.

$B=\frac{{\mu }_{0}I}{2\pi r}$ (1)

Here,

${\mu }_0$ is the constant.

$I$ is the amount of current flow in the wire.

$r$ is the distance of the wire from the origin.

From the given figure,

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

Substitute $\sqrt{a^2+z^2}$ for $r$ in the equation (1).

$B=\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+z^2}}$

The magnetic field component $B\sin \theta$ along $z$-axis cancel out with each other.

The resultant magnetic field along $y\text{-}$ axis is,

$\begin{array}{c}{B}_y=B\cos \theta +B\cos \theta \\ =2B\cos \theta \end{array}$

Substitute $\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+z^2}}$ for $B$ in the above equation.

$B_y=2\left(\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+z^2}}\right)\cos \theta$ . (2)

From the figure (2),

$\cos \theta =\frac{z}{\sqrt{a^2+z^2}}$

Substitute $\frac{a}{\sqrt{a^2+z^2}}$ for $\cos \theta$ in the equation (1).

$\begin{array}{c}{B}_y=2\left(\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+z^2}}\right)\left(\frac{z}{\sqrt{a^2+z^2}}\right)\\ =\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}\end{array}$ (3)

Conclusion:

Therefore, the value of magnetic field at origin is $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$.

(c)

To determine

The value of magnetic field at $y=0$, $z\to \infty$.

Answer to Problem 30.72CP

The value of magnetic field at $y=0$, $z\to \infty$ is $0$.

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

From the equation (3), the expression for magnetic field is,

$B_y=\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$

The value of magnetic field at $y=0$, $z\to \infty$ is,

$\begin{array}{c}\underset{z\to \infty }{\lim }{B}_y=\underset{z\to \infty }{\lim }\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}\\ {\left(B_y\right)}_{z=\infty }=\frac{{\mu }_{0}Iz}{\pi \left(a^2+{\infty }^2\right)}\\ =0\end{array}$

Conclusion:

Therefore, the value of magnetic field at $y=0$, $z\to \infty$ is $0$.

(d)

To determine

The magnetic field at points along the $z\text{-}$ axis as a function of $z$.

Answer to Problem 30.72CP

The magnetic field at points along the $z\text{-}$ axis as a function of $z$ is $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$.

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

From the calculated value in part (b), the magnetic field at points along the $z\text{-}$ axis as a function of $z$ is,

$B=\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$

Conclusion:

Therefore, the magnetic field at points along the $z\text{-}$ axis as a function of $z$ is $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$.

(e)

To determine

The distance along the positive $z\text{-}$ axis at which the magnetic field is maximum.

Answer to Problem 30.72CP

The magnetic field is maximum at $z=3.00\text{cm}$.

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

From the calculated value in part (b), the magnetic field at points along the $z\text{-}$ axis as a function of $z$ is,

$B=\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$

For maximum value of $B$.

$\frac{dB}{dz}=0$

Substitute $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$ for $B$ in the above equation.

$\begin{array}{c}\frac{d}{dz}\left\{\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}\right\}=0\\ \frac{{\mu }_{0}I}{\pi }\left(\frac{1}{\left(a^2+z^2\right)}-z\cdot \frac{2z}{{\left(a^2+z^2\right)}^2}\right)=0\\ \frac{{\mu }_{0}I}{\pi }\left(\frac{a^2-z^2}{a^2+z^2}\right)=0\\ a=z\end{array}$

Substitute $3.00\text{cm}$ for $a$ in the above equation.

$z=3.00\text{cm}$

Conclusion:

Therefore, the magnetic field is maximum at $z=3.00\text{cm}$.

(f)

To determine

The maximum value of magnetic field.

Explanation of Solution

Given info: The amount of current flow in infinitely long wire is $8.00\text{A}$ and the distance between the wires is $6.00\text{cm}$.

From the calculated value in part (b), the magnetic field at points along the $z\text{-}$ axis as a function of $z$ is,

$B=\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$

Substitute $a$ for $z$ in the above equation.

$\begin{array}{c}{B}_{\max }=\frac{{\mu }_{0}Ia}{\pi \left(a^2+a^2\right)}\\ =\frac{{\mu }_{0}I}{2\pi a}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_0$, $8.00\text{A}$ for $I$ and $3.00\text{cm}$ for $a$ in the above equation.

$\begin{array}{c}{B}_{\max }=\frac{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(8.00\text{A}\right)}{2\pi \left(3.00\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(8.00\text{A}\right)}{2\pi \left(3.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(8.00\text{A}\right)}{2\pi \left(0.03\text{m}\right)}\\ =5.33\times {10}^{-5}\text{T}\end{array}$

Simplify further,

$\begin{array}{c}{B}_{\max }=\left(5.33\times {10}^{-5}\text{T}\right)\left(\frac{1\text{μT}}{{10}^{-6}\text{T}}\right)\\ =53.3\text{μT}\end{array}$

Conclusion:

Therefore, the maximum value of magnetic field is $53.3\text{μT}$.