Chapter 30, Problem 20PQ

Two long, straight, parallel wires carry current as shown in Figure P30.18. If the currents are equal, find an expression for the magnetic field at point C. Use the indicated coordinate system to write your answer in component form.

FIGURE P30.18

To determine

Find the expression for the magnetic field at point C in component form

Answer to Problem 20PQ

The expression for net magnetic field at point C is $\frac{0.40{\mu }_{0}I}{\pi r}\text{T}\left(\widehat{i}\right)$.

Explanation of Solution

Redraw the figure P30.19 as shown below:

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

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

Here, $B$ is the magnetic field, $I$ is current, ${\mu }_0$ is permeability of free space and $r$ is the distance of the point from the wire.

The direction of the magnetic field at a point due to a current carrying wire is given by the right hand palm rule.

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

The current flowing in the wires is same and the point is at the equal distances from both the wire. Therefore, the vertical components of the magnetic fields will be same for both wires and they will cancel out each other.

Write the expression for the net magnetic field at point C refers to figure (a) above as.

    ${\overrightarrow{B}}_{net}=\left({\overrightarrow{B}}_1\sin \theta +{\overrightarrow{B}}_2\sin \theta \right)$                                                                             (II)

Here, ${\overrightarrow{B}}_{net}$ is net magnetic field, $B_1$ is magnetic field at point C due to wire carrying current $I_1$, $B_2$ is magnetic field at point C due to wire carrying current $I_2$ and $\theta$ is angle made by $\overrightarrow{B_1}$ from the vertical.

Apply Pythagoras theorem in triangle $\Delta I_{1}AC$ refer to figure (b) as.

    $\begin{array}{c}{r}_1=\sqrt{{\left(I_{1}A\right)}^2+{\left(AC\right)}^2}\\ =\sqrt{r^2+{\left(2r\right)}^2}\\ =\sqrt{r^2+4r^2}\\ =\sqrt{5r^2}\end{array}$

Simplify above equation as.

    $r_1=r\sqrt{5}$

Apply Pythagoras theorem in triangle $\Delta I_{2}AC$ refer to figure (b) as.

    $\begin{array}{c}{r}_2=\sqrt{{\left(I_{2}A\right)}^2+{\left(AC\right)}^2}\\ =\sqrt{r^2+{\left(2r\right)}^2}\\ =\sqrt{r^2+4r^2}\\ =\sqrt{5r^2}\end{array}$

Simplify above equation as.

    $r_2=r\sqrt{5}$

The net magnetic field due to the two wires is in the positive X-direction. The components of the field in the X-direction are calculated as below.

Substitute ${\overrightarrow{B}}_1\sin \theta$ for $B$, $r_1$ for $r$ and $I_1$ for $I$ in equation (I).

    ${\overrightarrow{B}}_1\sin \theta =\frac{{\mu }_0{I}_1}{2\pi r_1}\sin \theta$                                                                                      (III)

Here, ${\overrightarrow{B}}_1\sin \theta$ is the horizontal component of the magnetic field due to the wire carrying current $I_1$ and $r_1$ is distance of the point from the wire.

Substitute ${\overrightarrow{B}}_2\sin \theta$ for $B$, $r_2$ for $r$ and $I_2$ for $I$ in equation (I).

    ${\overrightarrow{B}}_2\sin \theta =\frac{{\mu }_0{I}_2}{2\pi r_2}\sin \theta$                                                                                     (IV)

Here, ${\overrightarrow{B}}_2\sin \theta$ is the horizontal component of the magnetic field due to the wire carrying current $I_2$ and $r_2$ is distance of the point from the wire.

Write the expression for the value of $\theta$ from figure (b) as.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{2r}{r}\right)\\ ={\tan }^{-1}\left(2\right)\\ \approx 63.5°\end{array}$

Substitute $r\sqrt{5}$ for $r_1$ and $I$ for $I_1$ in equation (III).

    ${\overrightarrow{B}}_1\sin \theta =\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \theta$

Substitute $r\sqrt{5}$ for $r_2$ and $I$ for $I_2$ in equation (IV).

    ${\overrightarrow{B}}_2\sin \theta =\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \theta$

Conclusion:

Substitute $\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \theta$ for ${\overrightarrow{B}}_1\sin \theta$, $\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \theta$ for ${\overrightarrow{B}}_2\sin \theta$ and $63.5°$ for $\theta$ in equation (II).

    $\begin{array}{c}{\overrightarrow{B}}_{net}=\left(\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \left(63.5°\right)+\frac{{\mu }_{0}I}{2\pi r\sqrt{5}}\sin \left(63.5°\right)\right)\left(\widehat{i}\right)\text{T}\\ =\left(\frac{{\mu }_{0}I}{\pi r\sqrt{5}}\sin \left(63.5°\right)\right)\left(\widehat{i}\right)\text{T}\\ =\left(0.40\right)\frac{{\mu }_{0}I}{\pi r}\left(\widehat{i}\right)\text{T}\end{array}$

Thus, the expression for net magnetic field at point C is $\frac{0.40{\mu }_{0}I}{\pi r}\text{T}\left(\widehat{i}\right)$.