91影视

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 $\stackrel{鈫�}{B}$is $胃=20.0掳$.

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

Formulae:

$B_{straight}=\frac{{渭}_{0}i}{4蟺R}$

$\tan 胃=\frac{B_y}{B_x}$

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

$B_{straight}=\frac{{渭}_{0}i}{4蟺R}$.

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

$B_1伪i_1$ and $B_2伪i_2$

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" $\tan 胃=\frac{B_y}{B_x}$.

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

Substitute the values and solve as:

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

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

$$\begin{gathered} 胃\text{'}=53.13掳-20.0掳 \\ 胃\text{'}=33.13掳 \end{gathered}$$

The final value of the current $i_1$is:

$\tan 胃\text{'}=\frac{i_2}{i_1}$

$i_1=\frac{i_2}{\tan 胃\text{'}}$

Substitute the values and solve as:

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