**Part A:** What is the magnetic field strength at point 1 in the figure?(B_1= ? T)

**Part B:** What is the magnetic field direction at point 1 in the figure? (out of the page or into the page?)

**Part C:** What is the magnetic field strength at point 2 in the figure?(B_2= ? T)

**Part D:** What is the magnetic field direction at point 2 in the figure?

**Part E:** What is the magnetic field strength at point 3 in the figure?(B_3= ? T)

**Part F:** What is the magnetic field direction at point 3 in the figure?

### Answer

**Part A:** From the Right-hand rule, the direction of the magnetic field at the point 1 (current points to the right) due to the top wire is out of the page and the direction of the magnetic field at the point 1 (current points to the left) due to the bottom wire is into the page. Thus, the net magnetic field at the point 1 die to both wires is,

```
\begin{aligned}
B_{1} =\left(\frac{\mu_{0} I_{\text {top }}}{2 \pi d_{1}}\right)_{\text {out }}+\left(\frac{\mu_{0} I_{\text {bottom }}}{2 \pi d_{2}}\right)_{i t} \\
=\frac{\mu_{0} I_{\text {top }}}{2 \pi d_{1}}-\frac{\mu_{0} I_{\text {bottom }}}{2 \pi d_{2}} \\
=\frac{\mu_{0}}{2 \pi}\left[\frac{I_{\text {top }}}{d_{1}}-\frac{I_{\text {bottom }}}{d_{2}}\right]
\end{aligned}
```

Here, `I_{\mathrm{meq}}`

is the current in the top wire, `I_{\mathrm{mom}}`

is the current in the bottom wire, `\mu_{0}`

is the permeability of the free space.

```
\begin{aligned}
B_{1} =\frac{\mu_{0}}{2 \pi}\left[\frac{I_{\operatorname{top}}}{d_{1}}-\frac{I_{\text {bottom }}}{d_{2}}\right] \\
=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)}{2 \pi}\left[\frac{10 \mathrm{~A}}{2.0 \mathrm{~cm}}-\frac{10 \mathrm{~A}}{6.0 \mathrm{~cm}}\right] \\
=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)}{2 \pi}\left[\frac{10 \mathrm{~A}}{2.0 \times 10^{-2} \mathrm{~m}}-\frac{10 \mathrm{~A}}{6.0 \times 10^{-2} \mathrm{~m}}\right] \\
=6.7 \times 10^{-5} \mathrm{~T}
\end{aligned}
```

**Part B:** The net magnetic field due to both wires at point 1 is directed out of the page.

**Part C:** From the Right-hand grip rule, the direction of the magnetic field at the point 2 due to the top wire (current points to the right) is into the page and the direction of the magnetic field at point 2 due to the bottom wire (current points to the left) is into the page.

Thus, the net magnetic field at the point 2 due to both wires is.

```
\begin{aligned}
B_{2} =\left(\frac{\mu_{0} I_{\operatorname{top}}}{2 \pi d_{1}}\right)_{\text {in }}+\left(\frac{\mu_{0} I_{\text {bottom }}}{2 \pi d_{2}}\right)_{\text {in }} \\
=-\frac{\mu_{0} I_{\text {top }}}{2 \pi d_{1}}-\frac{\mu_{0} I_{\text {bottom }}}{2 \pi d_{2}} \\
=-\frac{\mu_{0}}{2 \pi}\left(\frac{I_{\text {top }}}{d_{1}}+\frac{I_{\text {bortom }}}{d_{1}}\right)_{\text {in }} \\
=-\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)}{2 \pi}\left[\frac{10 \mathrm{~A}}{2.0 \mathrm{~cm}}-\frac{10 \mathrm{~A}}{2.0 \mathrm{~cm}}\right] \\
=-2.0 \times 10^{-4} \mathrm{~T}
\end{aligned}
```

Therefore, the magnitude of the magnetic field at the point 2 due to both wires is `2.0 \times 10^{-4} \mathrm{~T}`

. Here, the negative sign indicates the field is directed into the page.

**Part D:** The net magnetic field due to both wires at point 2 is directed into the page.

**Part E:** From the Right-hand grip rule, the direction of the magnetic field at the point `\mathbf{3}`

die to the top wire is into the page and the direction of the magnetic field at the point 3 due to the bottom wire is out of the page. Thus, the net magnetic field at the point `\mathbf{3}`

due to both wires is,

```
\begin{aligned}
B_{2} =\left(\frac{\mu_{0} I_{\mathrm{top}}}{2 \pi d_{1}}\right)_{\mathrm{in}}+\left(\frac{\mu_{0} I_{\mathrm{botom}}}{2 \pi d_{2}}\right)_{0 \mathrm{ut}} \\
=-\frac{\mu_{0} I_{\mathrm{top}}}{2 \pi d_{1}}+\frac{\mu_{0} I_{\text {bottom }}}{2 \pi d_{2}} \\
=\frac{\mu_{0}}{2 \pi}\left(-\frac{I_{\operatorname{top}}}{d_{1}}+\frac{I_{\text {botom }}}{d_{1}}\right)_{\text {in }} \\
=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)}{2 \pi}\left[-\frac{10 \mathrm{~A}}{6.0 \mathrm{~cm}}+\frac{10 \mathrm{~A}}{2.0 \mathrm{~cm}}\right] \\
=6.7 \times 10^{-5} \mathrm{~T}
\end{aligned}
```

Therefore, the magnitude of the magnetic field at the point 3 due to both wires is `6.7 \times 10^{-5} \mathrm{~T}`

. Here, the positive sign indicates the field is directed out of the page.

**Part F:**The net magnetic field due to both wires at point 2 is directed out of the page.