Question

A wire shaped in a regular hexagon of side 2 cm carries a current of 4 A. The magnetic field at the centre of the hexagon is.
 

• A. \( 4\sqrt{3} \times 10^{-5} \, \text{T} \)
• B. \( 8\sqrt{3} \times 10^{-5} \, \text{T} \)
• C. \( \sqrt{3} \times 10^{-5} \, \text{T} \)
• D. \( 6\sqrt{3} \times 10^{-5} \, \text{T} \)

Hint:

For a current-carrying wire shaped in a regular polygon, the magnetic field at the center is proportional to the number of sides of the polygon and inversely proportional to the radius.

Answer 1

The Correct Option is B

The magnetic field at the center of a regular polygon formed by a current-carrying wire is given by the formula: \[ B = \frac{\mu_0 I}{2 R} \times \text{number of sides} \] For a regular hexagon, the number of sides is 6, and the radius \( R \) is the distance from the center to a side. Given that the side length is 2 cm, we can use the geometry of the hexagon to find the radius. The magnetic field is calculated as: \[ B = \frac{4 \times 10^{-7} \times 4}{2 \times \left( \frac{2}{\sqrt{3}} \right)} = 8 \sqrt{3} \times 10^{-5} \, \text{T} \] Thus, the magnetic field at the center is \( 8 \sqrt{3} \times 10^{-5} \, \text{T} \).