Figure 29-45 shows two current segments. The lower segment carries a current of ${\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\text{A}}$and includes a semicircular arc with radius $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$, angle $\mathbf{180}\mathbf{°}$, and center point P. The upper segment carries current ${\mathit{i}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{2}{\mathit{i}}_{\mathbf{1}}$and includes a circular arc with radius $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$, angle $\mathbf{120}\mathbf{°}$, and the same center point P. What are the(a) magnitude and (b) direction of the net magnetic field at Pfor the indicated current directions? What are the (c)magnitude of if i₁ is reversed and (d) direction $\overrightarrow{\mathbf{B}}$ of if i₁ is reversed?

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${\theta }_1={180}^0=\pi rad$
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${\theta }_2={120}^0=\frac{2\pi }{3\text{rad}}$

Write the formula for the magnetic field:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}\mathbf{}\mathbf{\theta }\mathbf{}}{\mathbf{4}\mathbf{\pi }\mathbf{R}}$

The magnitude of the magnetic field is given by

$B=\frac{{\mu }_0{i}_1{\theta }_1}{4\pi r_1}\hat{k}-\frac{{\mu }_0{i}_2{\theta }_2}{4\pi r_2}$

Substitute the values and solve as:

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

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

The result of part a shows that the direction of the magnitude of the net magnetic field$\overrightarrow{B}$ 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

$B=-\frac{{\mu }_0{i}_1{\theta }_1}{4\pi r_1}\hat{k}-\frac{{\mu }_0{i}_2{\theta }_2}{4\pi r_2}$

Substitute the values and solve as:

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

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

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