Q27 - Fundamentals Of Physics

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 to rotate$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$ clockwise?

Step-by-Step Solution

Answer

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

Given

1. Currents flowing through the two long straight wires are  ${\text{i}}_1=30.0 mA$ and  ${\text{i}}_2=40.0 mA$.
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 $\tan \theta =\frac{B_y}{B_x}$.

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

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:

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