Solved Problem on Magnetic Field

Two equal circular loops E₁ and E₂, placed in perpendicular planes with coincident centers and radii R=5 π cm, carry currents i₁=3 A e i₂=4 A, as indicated in the figure. Assume \( \mu_{0}=4\pi \times 10^{-7}\,\frac{\text{T.m}}{\text{A}} \). Determine the vector of themagnetic field in the center O.

Problem data:

Radii of loops: R = 5 π cm;
Current in loop 1: i₁ = 3 A;
Current in loop 2: i₂ = 4 A;
Vacuum permeability: \( \mu_{0}=4\pi \times 10^{-7}\,\frac{\text{T.m}}{\text{A}} \).

Solution

First, we must convert the radii of the loops given in centimeters (cm) to meters (m) used in the International System of Units (S.I.).

\[ R=5\pi \;\cancel{\text{cm}}\times \frac{1\;\text{m}}{100\;\cancel{\text{cm}}}=0.05\pi\;\text{m}=5\times 10^{-2}\pi \;\text{m} \]

The field created by the current i₁, in the center of the loop E₁, can be obtained by applying the right-hand rule. Pointing the thumb in the direction of the current i₁ the other fingers will indicate the direction of the field, which in this case will be perpendicular to the plane of the loop and pointing into the page (Figure 1).
The magnitude of the magnetic field B is given by

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

Figure 1

For the magnetic field B₁, applying the expression (I), we have

\[ \begin{gather} B_{1}=\frac{\mu_{0}}{2}\frac{i_{1}}{r_{1}}\\ B_{1}=\frac{4\pi\times 10^{-7}}{2}\times \frac{3}{5\pi\times 10^{-2}}\\ B_{1}=\frac{12\times 10^{-7}}{10\times 10^{-2}}\\ B_{1}=12\times 10^{-7}\times 10\\ B_{1}=12\times 10^{-6}\;\text{T} \\ B_{1}=1.2\times 10^{-5}\;\text{T} \tag{II} \end{gather} \]

The field created by the current i₂, in the center of the loop E₂, can be obtained by applying the right-hand rule. Pointing the thumb in the direction of the current i₂ the other fingers will indicate the direction of the field, which in this case will be perpendicular to the plane of the loop and pointing downwards (Figure 2).

For the magnetic field B₂, applying the expression (I), we have

\[ \begin{gather} B_{2}=\frac{\mu_{0}}{2}\frac{i_{1}}{r_{1}}\\ B_{2}=\frac{4\pi\times 10^{-7}}{2}\times \frac{4}{5\pi\times 10^{-2}}\\ B_{2}=\frac{16\times 10^{-7}}{10\times 10^{-2}}\\ B_{2}=16\times 10^{-7}\times 10\\ B_{2}=16\times 10^{-6}\;\text{T}\\ B_{2}=1.6\times 10^{-5}\;\text{T} \tag{III} \end{gather} \]

Figure 2

The vectors magnetic fields, created in the center of the system by the currents in loops 1 and 2, are shown in Figure 3-A. Drawing these vectors in a three-dimensional coordinate system xyz (Figure 3-B), we obtain the resultant vector of the magnetic field B that makes an angle θ with the vertical z-direction.

Figure 3

We can calculate the magnitude of the resultant magnetic field using Pythagorean Theorem (Figure 3-C)

\[ \begin{gather} B^{2}=B_{1}^{2}+B_{2}^{2}\\ B^{2}=\left(1.2\times 10^{-5}\right)^{2}+\left(1.6\times 10^{-5}\right)^{2}\\ B^{2}=144\times 10^{-10}+2.56\times 10^{-10}\\ B^{2}=4.0\times 10^{-10}\\ B=\sqrt{4.0\times 10^{-10}\;}\\ B=2.0\times 10^{-5}\;\text{T} \end{gather} \]

The angle θ can be

\[ \begin{gather} \tan\theta =\frac{\text{opposite side}}{\text{adjacent side}}=\frac{B_{1}}{B_{2}}\\ \tan\theta =\frac{1.2\times 10^{-5}}{1.6\times 10^{-5}}\\ \tan\theta=0.75\\ \theta =\arctan(0.75)\simeq 37° \end{gather} \]

the magnetic field in the center of the system can be characterized by

Magnitude: B = 2.10⁻⁵ T ;
Direction: making an angle of approximately 37° with the z-axis pointing downwards. .

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .