A short magnet of moment `6.75 Am^(2)` produces a neutal point on its axis. If horizontal component of earth's magnetic field is `5xx10^(-5) "Wb"//m^(2)`, then the distance of the neutal point should be

Options

A. 10 cm
B. 20cm
C. 30cm
D. 40cm

Answer

To solve the problem, we need to find the distance of the neutral point from the short magnet. The neutral point occurs where the magnetic field due to the magnet equals the horizontal component of the Earth's magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic moment of the magnet, \( m = 6.75 \, \text{Am}^2 \)
- Horizontal component of the Earth's magnetic field, \( B_H = 5 \times 10^{-5} \, \text{Wb/m}^2 \)

2. **Understand the Magnetic Field Due to the Magnet:**
- The magnetic field \( B \) at a distance \( d \) from the center of a short magnet along its axis is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
- Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Set Up the Equation for the Neutral Point:**
- At the neutral point, the magnetic field due to the magnet \( B \) is equal to the horizontal component of the Earth's magnetic field \( B_H \):
\[ \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} = B_H \]

4. **Substituting the Values:**
- Substitute \( \mu_0 \) and \( m \) into the equation:
\[ \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 6.75}{d^3} = 5 \times 10^{-5} \]
- This simplifies to:
\[ 10^{-7} \cdot \frac{13.5}{d^3} = 5 \times 10^{-5} \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ \frac{13.5 \times 10^{-7}}{d^3} = 5 \times 10^{-5} \]
- Multiplying both sides by \( d^3 \) and dividing by \( 5 \times 10^{-5} \):
\[ d^3 = \frac{13.5 \times 10^{-7}}{5 \times 10^{-5}} \]

6. **Calculating \( d^3 \):**
- Calculate \( d^3 \):
\[ d^3 = \frac{13.5}{5} \times 10^{-2} = 2.7 \times 10^{-2} \]

7. **Finding \( d \):**
- Taking the cube root:
\[ d = (2.7 \times 10^{-2})^{1/3} \]
- This gives:
\[ d \approx 0.3 \, \text{m} \]

8. **Convert to Centimeters:**
- To convert meters to centimeters:
\[ d = 0.3 \, \text{m} \times 100 = 30 \, \text{cm} \]

### Final Answer:
The distance of the neutral point from the magnet is **30 cm**.

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