Due to a small magnet intensity at a distance `x` in the end on position is `9` Gauss. What will be the intensity at a distance `(x)/(2)` on broad side on position?

Options

A. 9 gauss
B. 4 gauss
C. 36 gauss
D. 4.5 gauss

Answer

To solve the problem, we need to determine the magnetic field intensity at a distance \( \frac{x}{2} \) on the broadside position of a small magnet, given that the intensity at a distance \( x \) on the end-on position is 9 Gauss.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Intensity**:
- The magnetic field intensity \( B \) due to a small magnet varies with distance. For a small magnet, the intensity at a distance \( r \) in the end-on position is given by:
\[ B_{\text{end-on}} = \frac{2m}{r^3} \]
- Where \( m \) is the magnetic moment of the magnet.

2. **Given Information**:
- At a distance \( x \) in the end-on position, the intensity \( B_{\text{end-on}} = 9 \) Gauss.
- Therefore, we can write:
\[ 9 = \frac{2m}{x^3} \quad \text{(1)} \]

3. **Finding the Magnetic Field Intensity at \( \frac{x}{2} \) on Broadside Position**:
- The intensity at a distance \( r \) in the broadside position is given by:
\[ B_{\text{broadside}} = \frac{m}{r^3} \]
- We need to find \( B_{\text{broadside}} \) at \( r = \frac{x}{2} \):
\[ B_{\text{broadside}} = \frac{m}{\left(\frac{x}{2}\right)^3} = \frac{m}{\frac{x^3}{8}} = \frac{8m}{x^3} \quad \text{(2)} \]

4. **Relating Equations (1) and (2)**:
- From equation (1), we have \( \frac{2m}{x^3} = 9 \), which implies:
\[ m = \frac{9x^3}{2} \quad \text{(3)} \]
- Substituting equation (3) into equation (2):
\[ B_{\text{broadside}} = \frac{8 \cdot \frac{9x^3}{2}}{x^3} = \frac{8 \cdot 9}{2} = 36 \text{ Gauss} \]

5. **Final Answer**:
- The intensity at a distance \( \frac{x}{2} \) on the broadside position is \( 36 \) Gauss.

### Summary:
The magnetic field intensity at a distance \( \frac{x}{2} \) on the broadside position is **36 Gauss**.