91Ó°ÊÓ

1. Current in the lower segment$i_1=0.40\text{A}$
2. Radius of the lower segment$r_1=5.0\text{cm}=0.05\text{m}$
3. Angle of the lower arc${\mathrm{θ}}_1={180}^0=\mathrm{π}rad$
4. Current in the upper segment$i_2=2i_1$
5. Radius of the lower segment$r_1=4.0\text{cm}=0.04\text{m}$
6. Angle of the lower arc${\mathrm{θ}}_2={120}^0=\frac{2\mathrm{π}}{3\text{rad}}$

Write the formula for the magnetic field:

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}\mathbf{}\mathbf{θ}\mathbf{}}{\mathbf{4}\mathbf{π}\mathbf{R}}$

The magnitude of the magnetic field is given by

$B=\frac{{\mathrm{μ}}_0{i}_1{\mathrm{θ}}_1}{4\mathrm{π}{r}_1}\hat{k}-\frac{{\mathrm{μ}}_0{i}_2{\mathrm{θ}}_2}{4\mathrm{π}{r}_2}$

Substitute the values and solve as:

$$\begin{gathered} B=\frac{{\mathrm{μ}}_0\left(0.40\text{A}\right)\left(\mathrm{π}\text{rad}\right)}{4\mathrm{π}\left(0.05\text{m}\right)}\hat{k}-\frac{{\mathrm{μ}}_0\left(0.80\text{A}\right)\left(\frac{2\mathrm{π}}{3}\text{rad}\right)}{4\mathrm{π}\left(0.04\text{m}\right)}\hat{k} \\ B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(8\mathrm{π}-\frac{40}{3}\mathrm{π}\right)\hat{k} \\ B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(-5.33\mathrm{π}\right)\hat{k} \\ B=-\left(1.7×{10}^{-6}\right)\hat{k} \end{gathered}$$

Therefore, the magnitude of the net magnetic field$\stackrel{→}{B}$ at P for the indicated current directions is $1.7×{10}^{-6}\text{T}$.

The result of part a shows that the direction of the magnitude of the net magnetic field$\stackrel{→}{B}$ at P for the indicated current directions is into the page, therefore, the direction is $-\hat{k}$

For the reverse direction of the current$i_1,$the magnitude of the magnetic field is given by

$B=-\frac{{\mathrm{μ}}_0{i}_1{\mathrm{θ}}_1}{4\mathrm{π}{r}_1}\hat{k}-\frac{{\mathrm{μ}}_0{i}_2{\mathrm{θ}}_2}{4\mathrm{π}{r}_2}$

Substitute the values and solve as:

$$\begin{gathered} B=-\frac{{\mathrm{μ}}_0\left(0.40\text{A}\right)\left(2\mathrm{π}\text{rad}\right)}{4\mathrm{π}\left(0.05\text{m}\right)}\hat{k}-\frac{{\mathrm{μ}}_0\left(0.80A\right)\left(\frac{2\mathrm{π}}{3}\text{rad}\right)}{4\mathrm{π}\left(0.04\text{m}\right)}\hat{k} \\ B=-\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(8\mathrm{π}-\frac{40}{3}\mathrm{π}\right)\hat{k} \\ B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(-21.33\mathrm{π}\right)\hat{k} \\ B=-\left(6.7×{10}^{-6}\right)\hat{k} \end{gathered}$$

Therefore, the magnitude of the magnetic field$\stackrel{\mathbf{→}}{\mathbf{B}}$ if$i_1$ is reversed is $6.7×{10}^{-6}\text{T}$.

The result of part c shows that the direction of the magnitude of the net magnetic field$\stackrel{→}{B}$ at P for the indicated current directions is into the page, therefore, the direction is $-\hat{k}$.