Chapter 30, Problem 72CP

(a)

To determine

The sketch of the magnetic field pattern in the $yz$ plane.

Answer to Problem 72CP

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

Explanation of Solution

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

Figure-(1)

The current in the infinitely long wires are in the negative $x$ direction, so by the right hand rule the magnetic field lines will be inside the sheet.

(b)

To determine

The value of the magnetic field at the origin.

Answer to Problem 72CP

The value of the magnetic field at the origin is zero.

Explanation of Solution

There is symmetry in the loop, so the contribution of each wire to the magnetic field at the origin will be same in magnitude but opposite in the direction. So the net magnetic field at the origin will be zero.

Therefore, the value of the magnetic field at the origin is zero.

(c)

To determine

The value of the magnetic field at $\left(y=0,z\to \infty \right)$.

Answer to Problem 72CP

The value of the magnetic field at $\left(y=0,z\to \infty \right)$ is zero.

Explanation of Solution

Write the expression for the magnetic field.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current in the wire, $a$ is the half distance between the two wires and $r$ is the radius of the loop.

Conclusion:

Substitute $\infty$ for $z$ in equation (I) to find $B$.

    $\begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+{\left(\infty \right)}^2}}\\ B=0\end{array}$

Therefore, the value of the magnetic field at $\left(y=0,z\to \infty \right)$ is zero.

(d)

To determine

The magnetic field at points along the $z$ axis as a function of $z$.

Answer to Problem 72CP

The magnetic field at points along the $z$ axis as a function of $z$ is $\overrightarrow{B}=\frac{32\times {10}^{-7}z}{9\times {10}^{-4}+z^2}\widehat{\text{j}}$.

Explanation of Solution

Write the equation of the magnetic field in $y$ direction.

    $\begin{array}{c}{B}_{\text{y}}=2\left(\frac{{\mu }_{0}I}{2\pi \sqrt{a^2+z^2}}\right)\sin \theta \\ =\frac{{\mu }_{0}I}{\pi \sqrt{a^2+z^2}}\left(\frac{z}{\sqrt{a^2+z^2}}\right)\\ =\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}\end{array}$                                                                                 (II)

So, this can be written as,

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

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $8.00\text{A}$ for $I$, and $3.00\text{cm}$ for $a$ in equation (III) to find $\overrightarrow{B}$.

  $\begin{array}{c}\overrightarrow{B}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(8.00\text{A}\right)z}{\pi \left({\left(3.00\text{cm}\right)}^2+z^2\right)}\widehat{\text{j}}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(8.00\text{A}\right)z}{\pi \left({\left(3.00\text{cm}\times \left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+z^2\right)}\widehat{\text{j}}\\ =\frac{32\times {10}^{-7}z}{9\times {10}^{-4}+z^2}\widehat{\text{j}}\end{array}$                                                                  (IV)

Therefore, the magnetic field at points along the $z$ axis as a function of $z$ is $\overrightarrow{B}=\frac{32\times {10}^{-7}z}{9\times {10}^{-4}+z^2}\widehat{\text{j}}$.

(e)

To determine

The distance along the positive $z$ axis where the magnetic field will be maximum.

Answer to Problem 72CP

The distance along the positive $z$ axis where the magnetic field will be maximum is $3.00\text{cm}$.

Explanation of Solution

Write the condition the maximum magnetic field.

    $\frac{d{B}_{\text{y}}}{dz}=0$                                                                                                                (V)

Substitute $\frac{{\mu }_{0}Iz}{\pi \left(a^2+z^2\right)}$ for $B_{\text{y}}$ in equation (V) to find $z$.

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

Conclusion:

Substitute $3.00\text{cm}$ for $a$ in equation (VI) to find $d$.

    $\begin{array}{l}z=a=d\\ d=3.00\text{cm}\end{array}$

Therefore, the distance along the positive $z$ axis where the magnetic field will be maximum will be $3.00\text{cm}$.

(f)

To determine

The maximum value of the magnetic field.

Answer to Problem 72CP

The maximum value of the magnetic field is $53.3\widehat{\text{j}}\mu \text{T}$.

Explanation of Solution

Conclusion:

Substitute $3.00\text{cm}$ for $z$ in equation (IV) to find $\overrightarrow{B}$.

  $\begin{array}{c}\overrightarrow{B}=\frac{\left(32\times {10}^{-7}\right)\left(3.00\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{\left(9\times {10}^{-4}\right)+{\left(3.00\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\widehat{j}\\ =53.3\times {10}^{-6}\widehat{\text{j}}\text{T}\times \left(\frac{{10}^6\mu \text{T}}{1\text{T}}\right)\\ =53.3\widehat{\text{j}}\mu \text{T}\end{array}$

Therefore, the maximum value of the magnetic field is $53.3\widehat{\text{j}}\mu \text{T}$.