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)/(a^(2))`
B. `(mu_(0))/(4pi)(6m)/(a^(2))`
C. `(mu_(0))/(4pi)(9m)/(a^(2))`
D. `(mu_(0))/(4pi)(m)/(2a^(2))`

Explanation

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

### Step 1: Understand the Setup
We have an equilateral triangle ABC with sides of length \( a \). At vertices B and C, we place north poles of pole strength \( m \), and at vertex A, we place a south pole of pole strength \( m \).

### Step 2: Determine the Center of the Triangle
The center of an equilateral triangle is the intersection of its medians. For triangle ABC, we denote the center as point O.

### Step 3: Calculate the Distance from the Center to Each Vertex
In an equilateral triangle, the distance from the center to any vertex can be calculated using the formula:
\[ AO = \frac{2}{3} \cdot AD \]
where \( AD \) is the median length. The median length \( AD \) can be calculated using the Pythagorean theorem:
\[ AD = \sqrt{a^2 - \left(\frac{a}{2}\right)^2} = \sqrt{a^2 - \frac{a^2}{4}} = \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}a}{2} \]
Thus,
\[ AO = \frac{2}{3} \cdot \frac{\sqrt{3}a}{2} = \frac{a\sqrt{3}}{3} \]

### Step 4: Calculate the Magnetic Field Due to Each Pole
The magnetic field \( B \) due to a magnetic pole of strength \( m \) at a distance \( r \) is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^2} \]
For the north poles at B and C, the distance \( r \) from O to B or C is \( AO = \frac{a\sqrt{3}}{3} \). Therefore, the magnetic field due to each north pole at point O is:
\[ B_N = \frac{\mu_0}{4\pi} \cdot \frac{m}{\left(\frac{a\sqrt{3}}{3}\right)^2} = \frac{\mu_0}{4\pi} \cdot \frac{m}{\frac{3a^2}{9}} = \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} \]

### Step 5: Calculate the Resultant Magnetic Field from the North Poles
Since both north poles repel a unit north pole placed at O, the angle between the magnetic fields \( B_N \) from poles B and C is \( 120^\circ \). The resultant magnetic field \( B_R \) can be calculated using the formula for the resultant of two vectors:
\[ B_R = \sqrt{B_N^2 + B_N^2 + 2B_NB_N\cos(120^\circ)} \]
Substituting \( \cos(120^\circ) = -\frac{1}{2} \):
\[ B_R = \sqrt{B_N^2 + B_N^2 - B_N^2} = \sqrt{B_N^2} = B_N \]
Thus, the resultant magnetic field due to the two north poles is:
\[ B_R = B_N = \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} \]

### Step 6: Calculate the Magnetic Field Due to the South Pole
The magnetic field \( B_S \) due to the south pole at A is directed towards the south pole and can be calculated similarly:
\[ B_S = \frac{\mu_0}{4\pi} \cdot \frac{m}{\left(\frac{a\sqrt{3}}{3}\right)^2} = \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} \]

### Step 7: Calculate the Total Magnetic Field at Point O
The total magnetic field at point O is the vector sum of \( B_R \) and \( B_S \). Since \( B_R \) is directed away from the north poles and \( B_S \) is directed towards the south pole, we add them:
\[ B_{total} = B_R + B_S = \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} + \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} = 2 \cdot \frac{3\mu_0}{4\pi} \cdot \frac{m}{a^2} = \frac{6\mu_0}{4\pi} \cdot \frac{m}{a^2} \]

### Final Answer
Thus, the intensity of the magnetic induction field strength at the center of the triangle is:
\[ B_{total} = \frac{6\mu_0}{4\pi} \cdot \frac{m}{a^2} \]