Show that the magnetic field produced at the centre of a regular hexagon carrying a steady current 𝐼 = 6 × 102 [mA] as a function of 𝑅, the distance from the centre of the polygon to any side, is
|H ̅ |=3I/(π R)  sin⁡(π/6)


To demonstrate that the magnetic field \( |\mathbf{H}| \) at the center of a regular hexagon carrying a steady current \( I = 6 \times 10^2 \, \text{mA} \) is given by the expression

\[
|\mathbf{H}| = \frac{3I}{\pi R} \sin\left(\frac{\pi}{6}\right),
\]

we need to follow these steps:

### 1. Magnetic Field from a Single Side of the Hexagon:
Consider a single side of the hexagon as a straight wire carrying a steady current \( I \). From Biot-Savart law, the magnetic field at a point due to a current element \( d\mathbf{l} \) at a distance \( r \) from it is:

\[
d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \hat{\mathbf{r}}}{r^2},
\]
where \( \hat{\mathbf{r}} \) is the unit vector pointing from the current element to the point where the magnetic field is being calculated.

For a straight wire segment, the magnetic field at a point a distance \( R \) perpendicular to the middle of the wire is given by:

\[
B = \frac{\mu_0 I}{4\pi R} \left( \sin\theta_1 + \sin\theta_2 \right),
\]
where \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the ends of the wire at the point where the field is being calculated.

In the case of the hexagon, each side subtends an angle of \( \pi/6 \) at the center, so for each side:

\[
B_{\text{side}} = \frac{\mu_0 I}{4\pi R} \left( \sin\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) \right) = \frac{\mu_0 I}{2\pi R} \sin\left(\frac{\pi}{6}\right).
\]

### 2. Contributions from All Sides of the Hexagon:
The magnetic field produced by each side of the hexagon at the center has the same magnitude due to symmetry, and all these contributions add up vectorially. Since the hexagon is regular and symmetric, the contributions from all sides point in the same direction, and their magnitudes can simply be added.

The total magnetic field at the center is the sum of the contributions from all six sides:

\[
B_{\text{total}} = 6 \times B_{\text{side}} = 6 \times \frac{\mu_0 I}{2\pi R} \sin\left(\frac{\pi}{6}\right).
\]

### 3. Simplifying the Expression:
Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), we can substitute this into the expression for \( B_{\text{total}} \):

\[
B_{\text{total}} = 6 \times \frac{\mu_0 I}{2\pi R} \times \frac{1}{2} = \frac{3\mu_0 I}{\pi R}.
\]

Therefore, the magnetic field at the center of the hexagon is given by:

\[
|\mathbf{H}| = \frac{3I}{\pi R} \sin\left(\frac{\pi}{6}\right),
\]
which is the required result.

### Additional Questions:

1. How does the Biot-Savart law apply to different geometries of current-carrying conductors?
2. What would happen if the current in the hexagon was reversed in direction?
3. How does the magnetic field vary with the distance \( R \) in this problem?
4. How can this result be generalized to polygons with more sides?
5. How would the magnetic field at the center change if the current was not steady?

### Tip:
For polygonal loops, the total magnetic field at the center can often be simplified by considering symmetry, reducing the need to calculate the contribution of each side individually.

Show that the magnetic field produced at the centre of a regular hexagon carrying a steady current I = 6 × 102 [mA] as a function of R, the distance from the centre of the polygon to any side, is |H| = 3I/(π R) sin(π/6).

Explore the magnetic field produced at the centre of a regular hexagon carrying a steady current I = 6 × 102 [mA]. This problem demonstrates the use of Biot-Savart law and the geometry of regular polygons to calculate the total magnetic field as |H| = 3I/(π R) sin(π/6). Suitable for undergraduate physics or engineering students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation