What is the magnetic field strength at point 1 in the figure?

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.