Chapter 30, Problem 83PQ

Two infinitely long current-carrying wires run parallel in the xy plane and are each a distance d = 11.0 cm from the y axis (Fig. P30.83). The current in both wires is I = 5.00 A in the negative y direction.

1. a. Draw a sketch of the magnetic field pattern in the xz plane due to the two wires. What is the magnitude of the magnetic field due to the two wires
2. b. at the origin and
3. c. as a function of z along the z axis, at x = y = 0?

FIGURE P30.83

To determine

The sketch of the magnetic field pattern in the x-z plane due to the two wires.

Answer to Problem 83PQ

The direction of the magnetic field pattern in x-z plane is shown in figure (a).

Explanation of Solution

The direction of the magnetic field for a current carrying wire is given by the Right Hand Palm rule.

According to the right hand palm rule, the thumb of the right hand points in the direction of the current flowing in the wire, the fingers point along the point at which the magnetic field is to be calculated then the palm faces towards the direction of the magnetic field.

Here, the direction of magnetic field due to both wires is in anticlockwise direction.

Conclusion:

Thus, the direction of the magnetic field pattern in x-z plane is shown in figure (a).

To determine

The magnitude of the magnetic field due to the two wires at the origin.

Answer to Problem 83PQ

The magnitude of the magnetic field due to the two wires at the origin in zero.

Explanation of Solution

The direction of the magnetic field due to first wire and the second wire is shown as.

Write the expression forthe magnetic field due to first wire.

  ${\overrightarrow{B}}_1\text{=}\frac{{\mu }_{0}I}{2\pi d}\left(\widehat{k}\right)$                                                                                                   (I)

Here ${\overrightarrow{B}}_1$ is the magnetic field due to first wire, $I$ is the current, $d$ is the distance of the origin from the wire.

Write the expression for the magnetic field due to second wire.

  ${\overrightarrow{B}}_2\text{=}\frac{{\mu }_{0}I}{2\pi d}\left(-\widehat{k}\right)$                                                                                               (II)

Here ${\overrightarrow{B}}_2$ is the magnetic field due to first wire, $I$ is the current and $d$ is the distance of the origin from the wire.

Write the expression for the net magnetic field at the origin as.

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

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

Conclusion:

Substitute $\left(\frac{{\mu }_{0}I}{2\pi d}\left(\widehat{k}\right)\right)$ for ${\overrightarrow{B}}_1$ and $\left(\frac{{\mu }_{0}I}{2\pi d}\left(-\widehat{k}\right)\right)$ for ${\overrightarrow{B}}_2$ in equation (III).

  $\begin{array}{c}{\overrightarrow{B}}_{\text{net}}=\left(\frac{{\mu }_{0}I}{2\pi d}\left(\widehat{k}\right)\right)+\left(\frac{{\mu }_{0}I}{2\pi d}\left(-\widehat{k}\right)\right)\\ =0\end{array}$

Hence, the net magnetic field at the origin will be zero.

To determine

The magnitude of the magnetic field due to two wire as a function of z along the z-axis.

Answer to Problem 83PQ

The magnetic field due to two wires as a  function of z along the Z-axis is $-\frac{\left(2\times {10}^{-6}\right)z}{\left(1.21\times {10}^{-2}\right)+z^2}\widehat{i}\text{T}$.

Explanation of Solution

The direction of the magnetic field for a current carrying wire is given by the Right Hand Palm rule.

The magnetic field due to both wires will be in negative X-direction at some point z above the origin.

Write the expression for the magnetic field due to first wire as.

  ${\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{2\pi r}\sin \theta \left(-\widehat{i}\right)$                                                                                     (IV)

Here, ${\overrightarrow{B}}_1$ is the magnetic field at point along Z-axis, ${\mu }_0$ is the permeability of free space, $I$ is the current in the wire, $\theta$ is the angle between r and x axis and $r$ is the distance between the wire and the point along the Z-axis.

The net magnetic field due to the second wire is same as that of the magnetic field due to the first wire.

Write the expression for the magnetic field due to second wire as.

  ${\overrightarrow{B}}_2=\frac{{\mu }_{0}I}{2\pi r}\sin \theta \left(-\widehat{i}\right)$                                                                                     (V)

Here, ${\overrightarrow{B}}_2$ is the magnetic field at point along Z-axis.

The distance of the point along the Z-axis from the wire is given by the Pythagoras theorem.

Write the expression for the distance $r$ from the wire as.

  $r=\sqrt{\left(z^2+d^2\right)}$                                                                                          (VI)

Here, $z$ is the vertical distance along the Z-axis and $d$ is the horizontal distance.

Write the expression for the angle made by the position vector with the horizontal as.

  $\sin \theta =\frac{z}{r}$

Substitute $\sqrt{\left(z^2+d^2\right)}$ for $r$ in above equation.

  $\sin \theta =\frac{z}{\sqrt{\left(z^2+d^2\right)}}$                                                                                    (VII)

Write the expression for the net magnetic field as.

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

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

Conclusion:

Substitute $\frac{z}{\sqrt{\left(z^2+d^2\right)}}$ for $\sin \theta$, $\sqrt{\left(z^2+d^2\right)}$ for $r$ in equation (IV).

  $\begin{array}{c}{\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{2\pi \sqrt{\left(z^2+d^2\right)}}\times \frac{z}{\sqrt{\left(z^2+d^2\right)}}(-\widehat{i})\\ =\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})\end{array}$

Substitute $\frac{z}{\sqrt{\left(z^2+d^2\right)}}$ for $\sin \theta$, $\sqrt{\left(z^2+d^2\right)}$ for $r$ in equation (V).

  $\begin{array}{c}{\overrightarrow{B}}_2=\frac{{\mu }_{0}I}{2\pi \sqrt{\left(z^2+d^2\right)}}\times \frac{z}{\sqrt{\left(z^2+d^2\right)}}(-\widehat{i})\\ =\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})\end{array}$

Substitute $\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})$ for ${\overrightarrow{B}}_1$ and $\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})$ for ${\overrightarrow{B}}_2$ in equation (VI).

  $\begin{array}{c}{\overrightarrow{B}}_{\text{net}}=\left(\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})\right)+\left(\frac{{\mu }_{0}Iz}{2\pi \left(z^2+d^2\right)}(-\widehat{i})\right)\\ =\frac{{\mu }_{0}Iz}{\pi \left(z^2+d^2\right)}(-\widehat{i})\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$, $5.00\text{A}$ for $I$ and $11.0\text{cm}$ for $d$ in above equation.

  $\begin{array}{c}{\overrightarrow{B}}_{\text{net}}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(5.00\text{A}\right)z}{\pi \left(z^2+{\left(11.0\text{cm}\right)}^2\right)}(-\widehat{i})\\ =-\frac{\left(2\times {10}^{-6}\right)z}{\left(z^2+{\left(11.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)}(-\widehat{i})\\ =-\frac{\left(2\times {10}^{-6}\right)z}{\left(1.21\times {10}^{-2}\right)+z^2}\widehat{i}\text{T}\end{array}$

Thus, the magnetic field due to two wires as a  function of z along the Z-axis is $-\frac{\left(2\times {10}^{-6}\right)z}{\left(1.21\times {10}^{-2}\right)+z^2}\widehat{i}\text{T}$.