Two north poles each of pole strength `m` and a south pole of pole strength `m` are placed at the three corners of an equilateral triangle of side a. The intensity of magnetic induction field strength at the centre of the triangle is

Options

A. `mu_0/(4pi) (m)/(da^2)`
B. `mu_0/(2pi) (3m)/(a^2)`
C. `mu_0/(4pi) (9m)/(a^2)`
D. `mu_0/(4pi) (m)/(2a^2)`

Answer

To solve the problem of finding the intensity of the magnetic induction field strength at the center of an equilateral triangle with two north poles and one south pole, we can follow these steps:

### Step 1: Understand the Configuration
We have an equilateral triangle with side length \( a \). The corners of the triangle have:
- North pole with pole strength \( m \) at vertex A,
- North pole with pole strength \( m \) at vertex B,
- South pole with pole strength \( m \) at vertex C.

### Step 2: Determine the Position of the Center
The center of an equilateral triangle is located at a distance of \( \frac{a}{\sqrt{3}} \) from each vertex. This distance will be used to calculate the magnetic field due to each pole.

### Step 3: Calculate the Magnetic Field due to Each Pole
The magnetic field \( B \) due to a magnetic pole is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^2} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( m \) is the pole strength,
- \( r \) is the distance from the pole to the point where the field is being calculated.

For our case, \( r = \frac{a}{\sqrt{3}} \).

### Step 4: Calculate the Magnetic Field from Each North Pole
For each north pole (A and B):
\[ B_A = B_B = \frac{\mu_0}{4\pi} \cdot \frac{m}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{\mu_0}{4\pi} \cdot \frac{m \cdot 3}{a^2} = \frac{3\mu_0 m}{4\pi a^2} \]

### Step 5: Determine the Direction of the Magnetic Fields
- The magnetic field due to a north pole points away from the pole. Therefore, the fields \( B_A \) and \( B_B \) will point outward from points A and B, respectively.
- The magnetic field due to the south pole (C) points towards the pole:
\[ B_C = \frac{\mu_0}{4\pi} \cdot \frac{m}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{3\mu_0 m}{4\pi a^2} \]
This field points towards the south pole at point C.

### Step 6: Resolve the Magnetic Fields into Components
- The magnetic fields \( B_A \) and \( B_B \) will have components along the vertical and horizontal axes.

For \( B_A \) and \( B_B \):
- The horizontal components will cancel each other out due to symmetry.
- The vertical components will add up.

The vertical component of \( B_A \) and \( B_B \) is:
\[ B_{Ay} = B_A \sin(60^\circ) = \frac{3\mu_0 m}{4\pi a^2} \cdot \frac{\sqrt{3}}{2} \]
\[ B_{By} = B_B \sin(60^\circ) = \frac{3\mu_0 m}{4\pi a^2} \cdot \frac{\sqrt{3}}{2} \]

### Step 7: Calculate the Total Vertical Component
The total vertical component from both north poles:
\[ B_{total\_y} = B_{Ay} + B_{By} = 2 \cdot \frac{3\mu_0 m \sqrt{3}}{8\pi a^2} = \frac{3\sqrt{3}\mu_0 m}{4\pi a^2} \]

### Step 8: Calculate the Contribution from the South Pole
The south pole contributes a downward magnetic field:
\[ B_C = \frac{3\mu_0 m}{4\pi a^2} \]

### Step 9: Combine the Contributions
The total magnetic field at the center is:
\[ B_{net} = B_{total\_y} - B_C = \frac{3\sqrt{3}\mu_0 m}{4\pi a^2} - \frac{3\mu_0 m}{4\pi a^2} \]
\[ B_{net} = \frac{3\mu_0 m}{4\pi a^2} \left(\sqrt{3} - 1\right) \]

### Final Result
The intensity of the magnetic induction field strength at the center of the triangle is:
\[ B_{net} = \frac{3\mu_0 m}{4\pi a^2} (\sqrt{3} - 1) \]