Two short bar magnets of magnetic moments `m` each are arranged at the opposite corners of a square of side d such that their centres coincide with thecorners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is

Options

A. `(mu_(0))/(4pi)(m)/(d^(3))`
B. `(mu_(0))/(4pi)(2m)/(d^(3))`
C. `(mu_(0))/(4pi)(m)/(2d^(3))`
D. `(mu_(0))/(4pi)(m^(3))/(2d^(3))`

Explanation

To solve the problem, we need to find the magnetic induction at one of the corners of the square formed by two bar magnets, each with a magnetic moment \( m \). The arrangement has the like poles facing each other. Let's denote the corners of the square as A, B, C, and D, where the magnets are placed at corners A and C, and we want to find the magnetic induction at corner B or D.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two bar magnets with magnetic moments \( m \) placed at opposite corners of a square of side \( d \).
- The magnets are aligned such that their like poles (e.g., north poles) are facing each other.

2. **Identify the Points**:
- Let’s denote the corners of the square as follows:
- A (where Magnet 1 is located)
- B (where we want to find the magnetic induction)
- C (where Magnet 2 is located)
- D (the other corner)
- We will calculate the magnetic induction at point B due to both magnets.

3. **Magnetic Field Due to Each Magnet**:
- The magnetic field \( B_1 \) at point B due to Magnet 1 (at corner A) can be calculated using the formula for the magnetic field along the axial line:
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
- The magnetic field \( B_2 \) at point B due to Magnet 2 (at corner C) can be calculated using the formula for the magnetic field along the equatorial line:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{m}{d^3} \]

4. **Direction of Magnetic Fields**:
- The magnetic field \( B_1 \) due to Magnet 1 at point B will point away from the magnet (towards the center of the square).
- The magnetic field \( B_2 \) due to Magnet 2 at point B will point towards the magnet (since it is an equatorial point).

5. **Resultant Magnetic Induction**:
- Since \( B_1 \) points away from Magnet 1 and \( B_2 \) points towards Magnet 2, the resultant magnetic induction \( B \) at point B is given by:
\[ B = B_1 - B_2 \]
- Substituting the values:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} - \frac{\mu_0}{4\pi} \cdot \frac{m}{d^3} \]
- Simplifying this expression:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m - m}{d^3} = \frac{\mu_0}{4\pi} \cdot \frac{m}{d^3} \]

6. **Final Result**:
- The magnetic induction at point B is:
\[ B = \frac{\mu_0 m}{4\pi d^3} \]

### Conclusion:
The magnetic induction at any of the other corners of the square is given by:
\[ \frac{\mu_0 m}{4\pi d^3} \]