Question

A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is: (Assume that the current is flowing in the clockwise direction)

• A. 3 x 10^-5 T

Answer 1

Answer: (A)

3 x 10^-5 T

Answer 2

To find the magnetic field at the centroid of an equilateral triangle carrying a current, we first calculate the magnetic field due to one side and then sum the contributions from all three sides.

1. Geometry of the Equilateral Triangle: Let the side length of the equilateral triangle be L. The centroid of an equilateral triangle is equidistant from each side. This perpendicular distance, let's call it a, can be found using the properties of an equilateral triangle. The height h of an equilateral triangle is h = L * sin(60°) = L * √3 / 2. The centroid is located at 1/3 of the height from the base. So, the perpendicular distance a from the centroid to any side is: $a = \frac{1}{3} h = \frac{1}{3} \left( \frac{L\sqrt{3}}{2} \right) = \frac{L}{2\sqrt{3}}$

Given L = 9 cm = 0.09 m. $a = \frac{0.09}{2\sqrt{3}} \text{ m}$

2. Magnetic Field due to a Finite Straight Wire: The magnetic field B at a point due to a finite straight current-carrying wire is given by: $B = \frac{\mu_0 I}{4\pi a} (\sin\theta_1 + \sin\theta_2)$ where:

• μ₀ is the permeability of free space ($4\pi \times 10^{-7} \text{ T·m/A}$).
• I is the current flowing through the wire.
• a is the perpendicular distance from the point to the wire.
• θ₁ and θ₂ are the angles made by the lines connecting the ends of the wire to the point, with the perpendicular line from the point to the wire.

For an equilateral triangle, consider one side. The centroid is the point. The perpendicular from the centroid to the midpoint of the side forms a right-angled triangle with half of the side and the line connecting the centroid to a vertex. The angle subtended by each half of the side at the centroid, with respect to the perpendicular, is 60 degrees. So, $\theta_1 = \theta_2 = 60^\circ$. $\sin\theta_1 = \sin\theta_2 = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

3. Magnetic Field due to One Side ($B_1$): Substitute the values into the formula for B: $B_1 = \frac{\mu_0 I}{4\pi a} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right)$ $B_1 = \frac{\mu_0 I}{4\pi a} (\sqrt{3})$ Now substitute the expression for a: $B_1 = \frac{\mu_0 I}{4\pi \left(\frac{L}{2\sqrt{3}}\right)} (\sqrt{3})$ $B_1 = \frac{\mu_0 I \cdot 2\sqrt{3}}{4\pi L} (\sqrt{3})$ $B_1 = \frac{\mu_0 I \cdot 2 \cdot 3}{4\pi L}$ $B_1 = \frac{6 \mu_0 I}{4\pi L} = \frac{3 \mu_0 I}{2\pi L}$

4. Total Magnetic Field at the Centroid: The current is flowing in the clockwise direction. Using the right-hand rule, the magnetic field produced by each side at the centroid will be directed into the plane of the triangle. Since the magnitudes are equal and directions are the same, the total magnetic field B_total is the sum of the magnetic fields due to each of the three sides. $B_{total} = 3 \times B_1$ $B_{total} = 3 \times \left( \frac{3 \mu_0 I}{2\pi L} \right)$ $B_{total} = \frac{9 \mu_0 I}{2\pi L}$

5. Substitute the Given Values:

• Current I = 1.5 A
• Side length L = 9 cm = 0.09 m
• Permeability of free space μ₀ = 4π × 10⁻⁷ T·m/A

$B_{total} = \frac{9 \times (4\pi \times 10^{-7}) \times 1.5}{2\pi \times 0.09}$ Cancel π from numerator and denominator: $B_{total} = \frac{9 \times 4 \times 10^{-7} \times 1.5}{2 \times 0.09}$ $B_{total} = \frac{36 \times 1.5 \times 10^{-7}}{0.18}$ $B_{total} = \frac{54 \times 10^{-7}}{0.18}$ To simplify $\frac{54}{0.18}$: $\frac{54}{0.18} = \frac{54}{\frac{18}{100}} = \frac{54 \times 100}{18} = 3 \times 100 = 300$ $B_{total} = 300 \times 10^{-7} \text{ T}$ $B_{total} = 3 \times 10^{-5} \text{ T}$

The final answer is $\boxed{3 \times 10^{-5} T}$.

The final answer is $\boxed{3 \times 10^{-5} T}$

Explanation of the solution: The magnetic field at the centroid of an equilateral triangle is the sum of the magnetic fields produced by each of its three sides. Due to symmetry and the clockwise current direction, the magnetic field contributed by each side at the centroid is equal in magnitude and points in the same direction (into the plane). The magnetic field due to a single straight wire segment is given by $B_1 = \frac{\mu_0 I}{4\pi a} (\sin\theta_1 + \sin\theta_2)$. For an equilateral triangle of side L, the perpendicular distance a from the centroid to a side is $a = \frac{L}{2\sqrt{3}}$. The angles $\theta_1$ and $\theta_2$ are both $60^\circ$. Substituting these values, the magnetic field due to one side is $B_1 = \frac{3 \mu_0 I}{2\pi L}$. The total magnetic field is $B_{total} = 3 \times B_1 = \frac{9 \mu_0 I}{2\pi L}$. Plugging in I = 1.5 A, L = 0.09 m, and μ₀ = 4π × 10⁻⁷ T·m/A, we get: $B_{total} = \frac{9 \times (4\pi \times 10^{-7}) \times 1.5}{2\pi \times 0.09} = \frac{54 \times 10^{-7}}{0.18} = 300 \times 10^{-7} = 3 \times 10^{-5} \text{ T}$.