Question

In Fig 29-55, two long straight wires (shown in cross section) carry currents ${\mathbf{i}}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mA}$and ${\mathbf{i}}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mA}$ directly out of the page. They are equal distances from the origin, where they set up a magnetic field. To what value must current be changed in order torotate$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$clockwise?

Short answer

The value of current is${\text{i}}_1=61.3mA$.

Step-by-step solution

Given

1. Currents flowing through the two long straight wires are ${\text{i}}_1=30.0mA$and ${\text{i}}_2=40.0mA$.
2. The rotation of net magnetic field $\overrightarrow{B}$is $\theta =20.0°$.

Determine the formula for the magnetic field

Use the concept of the magnetic force due to current in straight wires and trigonometry.

Formulae:

$B_{straight}=\frac{{\mu }_{0}i}{4\pi R}$

$\tan \theta =\frac{B_y}{B_x}$

Calculate the value to which current  must be changed in order to rotate 20.0° clockwise 

The magnetic field due to a current in the straight wire is

$B_{straight}=\frac{{\mu }_{0}i}{4\pi R}$.

The distances of the $B_1$and $B_2$are the same; hence they are directly proportional to $i_1$ and $i_2$respectively.

$B_1\alpha i_1$ and $B_2\alpha i_2$

According to the right hand rule, $B_2$is going along the y axis and $B_1$is going along x axis.

The angle of the net field is role="math" localid="1663076776348" $\tan \theta =\frac{B_y}{B_x}$.

$$\begin{gathered} \tan \theta =\frac{B_2}{B_1} \\ \theta ={\tan }^{-1}\left(\frac{i_2}{i_1}\right) \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{40.0\text{mA}}{30.0\text{mA}}\right) \\ \theta =53.13° \end{gathered}$$

In the problem, the net field rotation is$\theta \text{'}=\theta -20.0°$.

$$\begin{gathered} \theta \text{'}=53.13°-20.0° \\ \theta \text{'}=33.13° \end{gathered}$$

The final value of the current $i_1$is:

$\tan \theta \text{'}=\frac{i_2}{i_1}$

$i_1=\frac{i_2}{\tan \theta \text{'}}$

Substitute the values and solve as:

$i_1=\frac{40.0\text{mA}}{\tan \left(33.13°\right)}$