Question

A regular polygon of 6 sides is formed by bending a wire of length 4 \(\pi\) meter. If an electric current of \(4 \pi \sqrt3 \) A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be \(x × 10^{–7}\) T. The value of \(x\) is ____.

Answer 1

Correct Answer: 72

This problem requires us to calculate the magnetic field at the center of a regular hexagonal loop of wire. The total length of the wire and the current flowing through it are given.

Concept Used:

The magnetic field at a point due to a finite straight current-carrying wire is given by the Biot-Savart law application:

\[ B = \frac{\mu_0 I}{4\pi d} (\sin\theta_1 + \sin\theta_2) \]

where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \) T·m/A), \( I \) is the current, \( d \) is the perpendicular distance of the point from the wire, and \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the ends of the wire at that point.

For a regular n-sided polygon, the total magnetic field at the center is the sum of the magnetic fields due to each side. By symmetry, the field from each side is identical. Thus, the total field is:

\[ B_{\text{center}} = n \times B_{\text{one side}} \]

Step-by-Step Solution:

Step 1: List the given information and determine the side length of the hexagon.

Number of sides, \( n = 6 \).

Total length of the wire, \( L_{\text{total}} = 4\pi \) meters.

Current, \( I = 4\pi\sqrt{3} \) A.

The length of one side of the regular hexagon, \( a \), is the total length divided by the number of sides:

\[ a = \frac{L_{\text{total}}}{n} = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ meters} \]

Step 2: Determine the geometrical parameters for the magnetic field calculation.

A regular hexagon can be divided into 6 equilateral triangles with the center as a common vertex. The angle subtended by each side at the center is \( \frac{360^\circ}{6} = 60^\circ \). The angles \( \theta_1 \) and \( \theta_2 \) for the formula are half of this central angle.

\[ \theta_1 = \theta_2 = \frac{60^\circ}{2} = 30^\circ \]

The perpendicular distance, \( d \), from the center to a side is the altitude of one of these equilateral triangles.

\[ d = \frac{a\sqrt{3}}{2} = \left(\frac{2\pi}{3}\right) \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{3} = \frac{\pi}{\sqrt{3}} \text{ meters} \]

Step 3: Calculate the magnetic field at the center due to one side of the hexagon.

Using the formula for a finite wire:

\[ B_{\text{one side}} = \frac{\mu_0 I}{4\pi d} (\sin\theta_1 + \sin\theta_2) \]

Substitute the values:

\[ B_{\text{one side}} = \frac{(4\pi \times 10^{-7}) (4\pi\sqrt{3})}{4\pi (\frac{\pi}{\sqrt{3}})} (\sin 30^\circ + \sin 30^\circ) \]

Simplify the expression:

\[ B_{\text{one side}} = \frac{10^{-7} \times 4\pi\sqrt{3}}{\frac{\pi}{\sqrt{3}}} \left(\frac{1}{2} + \frac{1}{2}\right) \] \[ B_{\text{one side}} = 10^{-7} \times \frac{4\pi\sqrt{3} \times \sqrt{3}}{\pi} \times (1) \] \[ B_{\text{one side}} = 10^{-7} \times 4 \times 3 = 12 \times 10^{-7} \text{ T} \]

Step 4: Calculate the total magnetic field at the center of the hexagon.

The total magnetic field is the sum of the fields from all 6 sides. Since the field from each side points in the same direction (perpendicular to the plane of the polygon), we can simply multiply the field from one side by 6.

\[ B_{\text{center}} = n \times B_{\text{one side}} = 6 \times (12 \times 10^{-7} \text{ T}) \] \[ B_{\text{center}} = 72 \times 10^{-7} \text{ T} \]

Step 5: Determine the value of \( x \).

The problem states that the magnetic field is \( x \times 10^{-7} \) T. Comparing this with our calculated value:

\[ x \times 10^{-7} = 72 \times 10^{-7} \]

Therefore, the value of \( x \) is 72.

Answer 2

The magnetic field at the centre of a regular polygon with \(n\) sides is given by:

\[ B = \frac{\mu_0 I n}{4\pi r} \left(\sin 30^\circ + \sin 30^\circ\right) \]

Substituting values:

\[ B = 72 \times 10^{-7} \, \text{T} \]