91Ó°ÊÓ

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 $\stackrel{→}{B}$is$\mathrm{θ}=20.0°$.

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

Formulae:

$B_{straight}=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}R}$

$\tan \mathrm{θ}=\frac{B_y}{B_x}$

The value of current $i_1$:

The magnetic field due to a current in straight wire is

$B_{straight}=\frac{{\mathrm{μ}}_{0}i}{4\mathrm{π}R}$

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.

$B_1\mathrm{Î\pm }{i}_1$and

$B_2\mathrm{Î\pm }{i}_2$

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

The angle of the net field is

$\tan \mathrm{θ}=\frac{B_y}{B_x}$

$\tan \mathrm{θ}=\frac{B_2}{B_1}$

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

Substitute the values and solve as:

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

$\mathrm{θ}=53.13°$

In the problem, the net field rotation is

$\mathrm{θ}\text{'}=\mathrm{θ}-20.0°$

$\mathrm{θ}\text{'}=53.13°-20.0°$

$\mathrm{θ}\text{'}=33.13°$

The final value of the current is:

$\tan \mathrm{θ}\text{'}=\frac{i_2}{i_1}$

$i_1=\frac{i_2}{\tan \mathrm{θ}\text{'}}$

Substitute the values and solve as:

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

$i_1=61.3\text{mA}$