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 ${\mathbf{i}}_{\mathbf{1}}$be changed in order to rotate$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$clockwise?

i) Currents flowing through the two long straight wires are $i_1=30.0mA$and $i_2=40.0mA$

ii) The rotation of net magnetic field $\overrightarrow{B}$is$\theta =20.0°$.

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

Formulae:

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

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

The value of current $i_1$:

The magnetic field due to a current in 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 localid="1663143974221" $i_1$and $i_2$respectively.

$B_1\alpha i_1$and

$B_2\alpha i_2$

According to the right hand rule,is going to the y axis andis going along x axis.

The angle of the net field is

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

$\tan \theta =\frac{B_2}{B_1}$

$\theta ={\tan }^{-1}\left(\frac{i_2}{i_1}\right)$

Substitute the values and solve as:

$\theta ={\tan }^{-1}\left(\frac{40.0\text{mA}}{30.0\text{mA}}\right)$

$\theta =53.13°$

In the problem, the net field rotation is

$\theta \text{'}=\theta -20.0°$

$\theta \text{'}=53.13°-20.0°$

$\theta \text{'}=33.13°$

The final value of the current is:

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

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

Substitute the values and solve as:

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

$i_1=61.3\text{mA}$