A short bar magnet is arranged with its north pole pointing gergraphical north. It is found that the horizontal component of earth's magnetic induction `(B_(H))` is balaced by the magnetic induction of the magnet at a point which is at a distance of 20 cm from its centre .The magnetic moment of the magnet is `( if H = 4 xx 10^(-5) Wbm^(-2))`

Answer

To solve the problem, we need to find the magnetic moment \( m \) of a short bar magnet given that the horizontal component of the Earth's magnetic induction \( B_H \) is balanced by the magnetic induction of the magnet at a distance of 20 cm from its center. The value of \( B_H \) is given as \( 4 \times 10^{-5} \, \text{Wb/m}^2 \).

### Step-by-Step Solution:

1. **Understand the Formula:**
The horizontal component of the magnetic field due to a magnetic dipole (like a bar magnet) at a distance \( r \) is given by the formula:
\[ B_H = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( m \) is the magnetic moment,
- \( r \) is the distance from the center of the magnet.

2. **Substituting Known Values:**
We know:
- \( B_H = 4 \times 10^{-5} \, \text{Wb/m}^2 \)
- \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

Now, substituting these values into the formula:
\[ 4 \times 10^{-5} = \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{m}{(0.2)^3} \]

3. **Simplifying the Equation:**
The \( 4\pi \) terms cancel out:
\[ 4 \times 10^{-5} = \frac{10^{-7} \cdot m}{(0.2)^3} \]

Now calculate \( (0.2)^3 \):
\[ (0.2)^3 = 0.008 \, \text{m}^3 \]

Therefore, the equation becomes:
\[ 4 \times 10^{-5} = \frac{10^{-7} \cdot m}{0.008} \]

4. **Rearranging to Solve for \( m \):**
Multiply both sides by \( 0.008 \):
\[ 4 \times 10^{-5} \cdot 0.008 = 10^{-7} \cdot m \]

Calculate \( 4 \times 0.008 \):
\[ 4 \times 0.008 = 0.032 \]

Thus, we have:
\[ 0.032 \times 10^{-5} = 10^{-7} \cdot m \]

5. **Final Calculation:**
Convert \( 0.032 \times 10^{-5} \) to standard form:
\[ 0.032 \times 10^{-5} = 3.2 \times 10^{-7} \]

Now, divide both sides by \( 10^{-7} \):
\[ m = 3.2 \, \text{A m}^2 \]

### Final Answer:
The magnetic moment \( m \) of the magnet is \( 3.2 \, \text{A m}^2 \).