A current of i ampere is flowing in an equilateral triangle of side a. The magnetic induction at the centroid will be:
(1) \frac{\mu_0 i}{3\sqrt{3} \pi a}
(2) \frac{3\mu_0 i}{2\pi a}
(3) \frac{5\sqrt{2} \mu_0 i}{3\pi a}
(4) \frac{9\mu_0 i}{2\pi a}
Answer (1)
To determine the magnetic induction at the centroid of an equilateral triangle with a current flowing through its sides, we need to use the concept of the magnetic field due to a current-carrying conductor. This is an application of the Biot-Savart Law.
An equilateral triangle has three sides of equal length, and in this setup, each side of the triangle has a current i flowing through it. We are interested in finding the magnetic field at the centroid of the triangle.
The magnetic field due to a straight current-carrying wire at a perpendicular distance r from the wire is given by the formula: B = 2 π r μ 0 i where:
B is the magnetic field,
μ 0 is the permeability of free space,
i is the current,
r is the perpendicular distance from the wire to the point where the field is being calculated.
For an equilateral triangle, the centroid also serves as the point where the perpendicular bisectors of each side intersect. The distance from the centroid to any side of the triangle (perpendicular distance) is given by 2 3 a .
Using symmetry, the magnetic fields due to currents in the three sides of the triangle at the centroid can be vectorially added. Each side contributes equally to the magnetic field at the centroid, but because of the geometry, only components in one particular direction add constructively while others cancel out.
For each side of an equilateral triangle, the net magnetic field at the centroid is derived by calculating the component that does not cancel out due to symmetry. This results in: B = 3 × ( component of magnetic field from one side ) = 3 × ( 4 π × 2 3 a μ 0 i ) = 2 πa 3 μ 0 i
Therefore, the magnetic induction at the centroid of the triangle is given by option (2): 2 πa 3 μ 0 i
