Solved Problem on Magnetic Field
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In the figure, we have three conductors, A, B, and C, parallels. In the
wires flow the currents iA=10 A, iB=20 A and
iC=30 A, in the indicated directions, at a distance of
\( L=2\sqrt{3\;}\ \text{m} \)
from each other. Determine the vector of the magnetic field in the center O of the
triangle (centroid). Assume
\( \mu_{0}=4\pi \times 10^{-7}\frac{\text{T.m}}{\text{A}} \).
Problem data:
- Current on the wire A: iA = 10 A;
- Current on the wire B: iB = 20 A;
- Current on the wire C: iC = 30 A;
- Distance between wires: \( L=2\sqrt{3\;}\ \text{m} \);
- Vacuum permeability: \( \mu_{0}=4\pi \times 10^{-7}\frac{\text{T.m}}{\text{A}} \).
Solution
As the distances between the wires are the same, they occupy the vertices of an equilateral
triangle. The distance from the wires A, B, and C to the point O,
where we want the magnetic field, will be calculated based on Figure 2.
The altitude \( \overline{AM} \) is perpendicular to the side \( \overline{BC} \), and divides this side into two equal segments, so \( \overline{BM}=\frac{\overline{BC}}{2}=\frac{2\sqrt{3\;}}{2}=\sqrt{3\;}\;\text{m} \). The segment \( \overline{OB} \) divides the angle \( A\hat{B}C \) in half, so the angle \( O\hat{B}M \) is 30°, and the distance r to the center O will be \( r=\overline{OA}=\overline{OB}=\overline{OC} \).
The altitude \( \overline{AM} \) is perpendicular to the side \( \overline{BC} \), and divides this side into two equal segments, so \( \overline{BM}=\frac{\overline{BC}}{2}=\frac{2\sqrt{3\;}}{2}=\sqrt{3\;}\;\text{m} \). The segment \( \overline{OB} \) divides the angle \( A\hat{B}C \) in half, so the angle \( O\hat{B}M \) is 30°, and the distance r to the center O will be \( r=\overline{OA}=\overline{OB}=\overline{OC} \).
\[
\begin{gather}
\cos 30°=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{\overline{{BM}}}{r}\\
r=\frac{\overline{{BM}}}{\cos 30°}=\frac{\sqrt{3\;}}{\dfrac{\sqrt{3\;}}{2}}=\cancel{\sqrt{3\;}}\times \frac{2}{\cancel{\sqrt{3\;}}}=2\;\text{m}
\end{gather}
\]
The magnitude of the magnetic field of a wire is calculated by the expression
\[
\begin{gather}
\bbox[#99CCFF,10px]
{B=\frac{\mu _{0}}{2\pi}\frac{i}{r}} \tag{I}
\end{gather}
\]
The vector of magnetic field
\( {\vec{B}}_{A} \),
due to the current through the wire A, will be given by the right-hand rule, will be
tangent to the line of magnetic induction at point O and perpendicular to the segment
\( \overline{OA} \)
(Figure 3), applying the expression (I)
\[
\begin{gather}
B_{A}=\frac{\mu_{0}}{2\pi}\frac{i_{A}}{r}\\
B_{A}=\frac{\cancel{4}\cancel{\pi} \times 10^{-7}}{\cancel{2}\cancel{\pi}}\times \frac{10}{\cancel{2}}\\
B_{A}=10\times 10^{-7}\;\text{T}
\end{gather}
\]
Figure 3
The vector of magnetic field
\( {\vec{B}}_{B} \),
due to the current through the wire B, will be given by the right-hand rule, will be
tangent to the line of magnetic induction at point O and perpendicular to the segment
\( \overline{OB} \)
(Figure 4). The angle
\( O\hat{B}C \)
is equal to 30°, this angle and angle α are alternates angles, then α is also 30°.
The angle β will be
\[
\begin{gather}
\alpha +90°+\beta=180°\\
30°+90°+\beta =180°\\\
\beta=180°-120°\\\
\beta =60°
\end{gather}
\]
Applying the expression (I)
\[
\begin{gather}
B_{B}=\frac{\mu_{0}}{2\pi}\frac{i_{B}}{r}\\
B_{B}=\frac{\cancel{4}\cancel{\pi} \times 10^{-7}}{\cancel{2}\cancel{\pi}}\frac{20}{\cancel{2}}\\
B_{B}=20\times 10^{-7}\;\text{T}
\end{gather}
\]
The vector of magnetic field
\( {\vec{B}}_{C} \),
due to the current through the wire C, will be given by the right-hand rule, will be
tangent to the line of magnetic induction at point O and perpendicular to the segment
\( \overline{OC} \)
(Figure 5). The angle
\( O\hat{C}B \)
is equal to 30 °, this angle and angle α are alternates angles, then α is also 30°.
The angle β will be
\[
\begin{gather}
\alpha +\beta =90°\\
30°+\beta=90°\\
\beta =90°-30°\\
\beta=60°
\end{gather}
\]
it is the angle between the vector
\( {\vec{B}}_{C} \)
and the horizontal line (Figure 5).
Applying the expression (I)
\[
\begin{gather}
B_{B}=\frac{\mu_{0}}{2\pi}\frac{i_{B}}{r}\\
B_{B}=\frac{\cancel{4}\cancel{\pi} \times 10^{-7}}{\cancel{2}\cancel{\pi}}\frac{30}{\cancel{2}}\\
B_{B}=30\times 10^{-7}\;\text{T}
\end{gather}
\]
The resultant vector of the magnetic field will be calculated by the sum
\[
\vec{B}={\vec{B}}_{A}+{\vec{B}}_{B}+{\vec{B}}_{C}
\]
drawing the vectors
\( {\vec{B}}_{A} \),
\( {\vec{B}}_{B} \)
and
\( {\vec{B}}_{C} \)
in a coordinated system xy and calculating its components in x and y
directions (Figure 6)
Figure 6
Direction x:
\[
\begin{gather}
B_{Ax}=B_{A}\cos 0°=10\times 10^{-7}\times 1=10\times 10^{-7}\;\text{T}\\[5pt]
B_{BX}=B_{B}\cos (-60°)=B_{B}\cos 60°=20\times 10^{-7}\times \frac{1}{2}=10\times 10^{-7}\;\text{T}\\[5pt]
B_{Cx}=B_{C}\cos 60°=30\times 10^{-7}\times \frac{1}{2}=15\times 10^{-7}\;\text{T}
\end{gather}
\]
The magnitude of resultant vector in the x-direction will be
\[
\begin{gather}
B_{x}=B_{Ax}+B_{BX}+B_{Cx}\\
B_{x}=10\times 10^{-7}+10\times 10^{-7}+15\times 10^{-7}\\
B_{x}=35\times 10^{-7}\;\text{T}
\end{gather}
\]
Direction y:
\[
\begin{gather}
B_{Ay}=B_{A}\sin 0=10\times 10^{-7}\times 0=0\\[5pt]
B_{By}=B_{B}\sin(-60°)=-B_{B}\sin 60°=-20\times 10^{-7}\times \frac{\sqrt{3\;}}{2}=-10\sqrt{3\;}\times 10^{-7}\;\text{T}\\[5pt]
B_{Cy}=B_{C}\sin 60°=30\times 10^{-7}\times \frac{\sqrt{3\;}}{2}=15\sqrt{3\;}\times 10^{-7}\;\text{T}
\end{gather}
\]
The magnitude of resultant vector in the y-direction will be
\[
\begin{gather}
B_{y}=B_{Ay}+B_{By}+B_{Cy}\\
B_{y}=0-10\sqrt{3\;}\times 10^{-7}+15\sqrt{3\;}\times 10^{-7}\\
B_{y}=5\sqrt{3\;}\times 10^{-7}\;\text{T}
\end{gather}
\]
The magnitude of resultant vector
\( \vec{B} \)
will be calculated by applying the Pythagorean Theorem (Figure 7-A)
\[
\begin{gather}
B^{2}=B_{x}^{2}+B_{y}^{2}\\
B^{2}=\left(35\times 10^{-7}\right)^{2}+\left(5\sqrt{3\;}\times 10^{-7}\right)^{2}\\
B^{2}=\left(1225\times 10^{-14}\right)+\left(25\times 3\times 10^{-14}\right)\\
B^{2}=(1225+75)\times 10^{-14}\\
B=\sqrt{1300\times 10^{-14}\;}\\
B=3.6\times 10^{-6}\;\text{T}
\end{gather}
\]
Vector \( \vec{B} \) will form with the axis an angle θ given by
\[
\begin{gather}
\tan \theta =\frac{\text{opposite side}}{\text{adjacent side}}=\frac{B_{y}}{B_{x}}\\
\tan \theta=\frac{5\sqrt{3\;}.\cancel{10^{-7}}}{35.\cancel{10^{-7}}}\\
\tan \theta \simeq0,25\\
\theta =\arctan(0.25)\simeq 14°
\end{gather}
\]
The resultant vector is shown in Figure 7-B
Magnitude: 3.6\times 10−7 T ;
Direction: making an angle of 14º with horizontal pointing to the right. .
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .