# What is the magnetic field B? at point a in the figure?(Figure 1) Give your answer as vector.

Assume I=11A

Part A: What is the magnetic field B? at point a in the figure?(Figure 1) Give your answer as vector.

Part B: What is the magnetic field B? at point b in the figure? Give your answer as vector.

Part C: What is the magnetic field B? at point c in the figure? Give your answer as vector

(Part A). The magnetic field at the point a due to the top and bottom long wires is,

\vec{B}_{a}=\vec{B}_{\text {top }}+\vec{B}_{\text {bottom }}

The magnetic field at point a due to the top wire is,

B_{\text {top }}= \frac{\mu_{0} I}{2 \pi d}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi \sqrt{\left(1.0 \times 10^{-2} \mathrm{~m}\right)^{2}+\left(1.0 \times 10^{-2} \mathrm{~m}\right)^{2}}}

=1.5556 \times 10^{-4} \mathrm{~T}

The magnetic field at point $a$ due to thebottom wire is,

\begin{aligned}
B_{\text {bottom }} &=\frac{H_{0} I}{2 \pi d^{\prime}}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi \sqrt{\left(-1.0 \times 10^{-2} \mathrm{~m}\right)^{2}+\left(-1.0 \times 10^{-2} \mathrm{~m}\right)^{2}}} \\
&=1.5556 \times 10^{-4} \mathrm{~T}
\end{aligned}

From the figure, the total magnetic field at the point a is,

\begin{aligned}
\vec{B}=\left(B_{\operatorname{top}} \cos 45^{\circ}\right) \hat{i}-\left(B_{\operatorname{top}} \sin 45^{\circ}\right) \hat{j}+\left(B_{\text {Bottom }} \cos 45^{\circ}\right) \hat{i}+\left(B_{\text {Bottom }} \sin 45^{\circ}\right) \hat{j} \\
=2 B_{\operatorname{top}} \cos 45^{\circ} \hat{i} \\
=2\left(1.5556 \times 10^{-4} \mathrm{~T}\right) \cos 45^{\circ} \hat{i}=\left(2.2 \times 10^{-4} \mathrm{~T}\right) \hat{i}
\end{aligned}

(Part B). The magnetic field at point $b$ due to the top wire is,

B_{\text {top }}=\frac{\mu_{0} I}{2 \pi d}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi(0.01 \mathrm{~m})}=2.2 \times 10^{-4} \mathrm{~T}

The magnetic field at point $b$ due to thebottom wire is,

B_{\text {bottom }}=\frac{\mu_{0} I}{2 \pi d^{\prime}}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi(0.01 \mathrm{~m})}=2.2 \times 10^{-4} \mathrm{~T}

From the figure, the total magnetic field at the point b is,

\vec{B}=\left(B_{\mathrm{top}}\right) \hat{i}+\left(B_{\mathrm{bottom}}\right) \hat{i}

=2 B_{\mathrm{top}} \cos 45^{\circ} \hat{i}=2\left(2.2 \times 10^{-4} \mathrm{~T}\right) \hat{i}=\left(4.4 \times 10^{-4} \mathrm{~T}\right) \hat{i}

(Part C). The magnetic fiel $\mathrm{d}$ at point $c$ due to the top wire is,

\begin{aligned}
B_{\text {top }}= \frac{\mu_{0} I}{2 \pi d}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi \sqrt{\left(1.0 \times 10^{-2} \mathrm{~m}\right)^{2}+\left(1.0 \times 10^{-2} \mathrm{~m}\right)^{2}}} \\
=1.5556 \times 10^{-4} \mathrm{~T}
\end{aligned}

The magnetic field at point $c$ due to thebottom wire is,

\begin{aligned}
B_{\mathrm{bottom}} =\frac{H_{0} I}{2 \pi d^{\prime}}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(11 \mathrm{~A})}{2 \pi \sqrt{\left(-1.0 \times 10^{-2} \mathrm{~m}\right)^{2}+\left(-1.0 \times 10^{-2} \mathrm{~m}\right)^{2}}} \\
=1.5556 \times 10^{-4} \mathrm{~T}
\end{aligned}

From the figure, the total magnetic field at the point c is,

\begin{aligned}
\vec{B}=\left(B_{\operatorname{top}} \cos 45^{\circ}\right) \hat{i}-\left(B_{\operatorname{top}} \sin 45^{\circ}\right) \hat{j}+\left(B_{\text {Bottom }} \cos 45^{\circ}\right) \hat{i}+\left(B_{\text {Bottom }} \sin 45^{\circ}\right) \hat{j} \\
=2 B_{\operatorname{top}} \cos 45^{\circ} \hat{i}=2\left(1.5556 \times 10^{-4} \mathrm{~T}\right) \cos 45^{\circ} \hat{i}=\left(2.2 \times 10^{-4} \mathrm{~T}\right) \hat{i}
\end{aligned}

### Raymond Puzio

Raymond Puzio has a PhD in Physics from Yale University. I have been creating PlanetPhysics with Aaron Krowne and Ben Loftin since 2005.