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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)

I have got to dH ̅=I/4π  dl/R^2 but now am stuck

You're on the right track with using the Biot-Savart law to find the magnetic field at the center of the regular hexagon due to a steady current, but let's refine the approach to proceed from where you're stuck.

### Step-by-Step Solution:

#### 1. **Magnetic Field Contribution from a Small Element (Biot-Savart Law)**:
The Biot-Savart law gives the magnetic field contribution \( d\mathbf{H} \) at a point due to a current element \( dl \):
\[
d\mathbf{H} = \frac{I}{4\pi} \frac{d\mathbf{l} \times \mathbf{r}}{r^3}
\]
where:
- \( I \) is the current in the wire,
- \( d\mathbf{l} \) is the infinitesimal length of the current-carrying wire,
- \( \mathbf{r} \) is the position vector from the current element to the point where the magnetic field is being calculated (in this case, the center of the hexagon),
- \( r = R \) is the distance from the center of the hexagon to any side of the hexagon (constant for all sides).

For a regular hexagon, we can use symmetry and integrate the magnetic field contributions from all six sides.

#### 2. **Geometry of the Hexagon**:
In a regular hexagon:
- The distance from the center to each side is \( R \),
- The total length of each side is \( L = 2R \tan(\pi/6) = R\sqrt{3} \).

#### 3. **Symmetry and Integration**:
Due to the hexagon's symmetry, the contributions to the magnetic field from each of the six sides will be identical in magnitude, and their directions will sum constructively due to the geometry. Each side contributes an angle \( \theta = \pi/6 \).

You are already at:
\[
d\mathbf{H} = \frac{I}{4\pi} \frac{dl}{R^2}
\]
But because of symmetry, the magnetic field from each side will have a factor of \( \sin(\pi/6) \), as only the perpendicular component contributes to the field at the center.

#### 4. **Summing the Contributions**:
Since the current is the same in all segments, the total magnetic field is just the sum of the fields from each of the six sides. The contributions add up because the vectors from each side align symmetrically towards the center.

Now integrate over the length of one side:
\[
H_{\text{side}} = \int \frac{I}{4\pi} \frac{dl \sin(\pi/6)}{R^2}
\]
This simplifies as \( \sin(\pi/6) = 1/2 \).

Then summing for all 6 sides, the total magnetic field is:
\[
H = 6 \cdot \frac{I}{4\pi} \cdot \frac{\sin(\pi/6)}{R} = 6 \cdot \frac{I}{4\pi} \cdot \frac{1}{2R}
\]

Simplifying:
\[
H = \frac{3I}{\pi R}
\]

Thus, the magnetic field at the center of the hexagon is:
\[
|\mathbf{H}| = \frac{3I}{\pi R} \sin\left(\frac{\pi}{6}\right)
\]

This matches the required result.

---

Would you like further clarification or additional details on any step?

Here are 5 related questions to consider:
1. How does the magnetic field change if the current \( I \) is doubled?
2. What happens to the magnetic field if the number of sides in the polygon increases?
3. Can this method be generalized for a polygon with \( N \) sides?
4. How would the magnetic field change if the current flows in the opposite direction?
5. What if the current-carrying shape was not planar (e.g., 3D coil)?

**Tip**: Symmetry is a powerful tool in physics problems—it often allows you to simplify the calculations and focus on essential components.

Show that the magnetic field produced at the centre of a regular hexagon carrying a steady current 𝐼 = 6 × 10^2 [mA] as a function of 𝑅, the distance from the centre of the polygon to any side, is |H| = 3I/(πR) sin(π/6). I have got to dH = I/4π dl/R^2 but now am stuck.

This solution demonstrates how to calculate the magnetic field at the center of a regular hexagon carrying a steady current of 6 × 10^2 mA using the Biot-Savart law. The calculation involves symmetry and integration, yielding the final result H = 3I/(πR) * sin(π/6). Suitable for undergraduate physics students.

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