Solved Problem on Magnetic Field

Three straight and long wire form the equilateral triangle of the figure without touching themselves. The circle in the triangle has center and radius r = 10 cm. The currents in conductors have the directions indicated in the figure, and are equal to i₁ = i₂ = 10 A and i₃ = 5 A. Determine the resultant magnetic field vector in C \( \left(\mu_{0}=4\pi \times 10^{-7}\;\frac{\text{T.m}}{\text{A}}\right) \).

Problem data:

Circle radius: r = 10 cm;
Current on wire 1: i₁ = 10 A;
Current on wire 2: i₂ = 10 A;
Current on wire 3: i₃ = 5 A..

Problem diagram:

The distance from the center to the wires are all equal to r (Figure 1-A), we will choose a reference frame in the center of the triangle, with a positive directon pointing out (leaving the sheet) and perpendicular to the triangle plane (Figure 1-B).

Figure 1

Solution

First we convert the radius of the circle given into centimeters (cm) to meters (m) used in the International System of Units (SI)

\[ \begin{gather} r=10\;\cancel{\text{cm}}\times {\frac{1\;\text{m}}{100\;\cancel{\text{cm}}}}=0.1\;\text{m} \end{gather} \]

The magnetic field of a straight wire is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {B=\frac{\mu_{0}}{2\pi }\frac{i}{r}} \tag{I} \end{gather} \]

Applying the Right Hand Rule to the wire where the i₁ current passes, placing the thumb in the direction of the current the other fingers indicate that the magnetic field is towards inside (it is entering the sheet - Figure 2). Applying the equation (I)

\[ \begin{gather} B_{1}=\frac{4\pi \times 10^{-7}}{2\pi}\times\frac{10}{0.1}\\[5pt] B_{1}=2\times 10^{-7}\times 100\\[5pt] B_{1}=2\times10^{-5}\;\text{T} \end{gather} \]

Figure 2

Applying the Right Hand Rule to the wire where the i₂ current passes, the magnetic field is directed inside (it is entering the sheet - Figure 3).

\[ \begin{gather} B_{2}=\frac{4\pi \times 10^{-7}}{2\pi}\times\frac{10}{0.1}\\[5pt] B_{2}=2\times 10^{-7}\times 100\\[5pt] B_{2}=2\times10^{-5}\;\text{T} \end{gather} \]

Figure 3

Applying the Right Hand Rule to the wire where the i₃ current passes, the magnetic field is meaningful (it is entering the sheet - Figure 4). Applying the equation (I)

\[ \begin{gather} B_{3}=\frac{4\pi \times 10^{-7}}{2\pi}\times\frac{5}{0.1}\\[5pt] B_{3}=2\times 10^{-7}\times 50\\[5pt] B_{3}=1\times10^{-5}\;\text{T} \end{gather} \]

Figure 4

The magnetic field vector will be given by the sum of the three vectors

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{B}={\vec{B}}_{1}+{\vec{B}}_{2}+{\vec{B}}_{3}} \end{gather} \]

the magnitude will be

\[ \begin{gather} B=-B_{1}+B_{2}+B_{3}\\[5pt] B=-2\times 10^{-5}+2\times10^{-5}+1\times 10^{-5}\\[5pt] B=1\times 10^{-5} \end{gather} \]

The resultant vector is (Figure 5)

Magnitude: 1× 10⁻⁵ T;

Direction: perpendicular to the triangle plane and pointing out.

Figure 5

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