A small magnet of dipole moment `M` is kept on the arm of a deflection magnetometer set in tan `A` position at a distance of `0.2 m`. If the deflection is `60^(@)`, find the value of `P (B_(H)=0.4xx10^(-4)T)`.

Options

A. `2.77 "A m"^2`
B. `8 "A m"^2`
C. `0.2 "A m"^2`
D. `0.9 "A m"^2`

Answer

To solve the problem step by step, we will follow the outlined process in the video transcript.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Distance \( d = 0.2 \, \text{m} \)
- Deflection angle \( A = 60^\circ \)
- Magnetic field \( B_H = 0.4 \times 10^{-4} \, \text{T} \)

2. **Use the Formula for a Small Magnet Dipole in Tan A Position:**
The formula relating the dipole moment \( P \), the magnetic field \( B_H \), and the distance \( d \) is given by:
\[ \frac{\mu_0}{4\pi} \cdot \frac{2P}{d^3} = B_H \tan A \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substituting Known Values:**
Substitute the known values into the equation:
\[ \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2P}{(0.2)^3} = 0.4 \times 10^{-4} \cdot \tan(60^\circ) \]
Simplifying gives:
\[ \frac{10^{-7} \cdot 2P}{(0.2)^3} = 0.4 \times 10^{-4} \cdot \sqrt{3} \]

4. **Calculate \( (0.2)^3 \):**
\[ (0.2)^3 = 0.008 \, \text{m}^3 \]

5. **Rearranging the Equation:**
Now, we can rearrange the equation to solve for \( P \):
\[ 2P = \frac{0.4 \times 10^{-4} \cdot \sqrt{3} \cdot (0.2)^3}{10^{-7}} \]
\[ P = \frac{0.4 \times 10^{-4} \cdot \sqrt{3} \cdot 0.008}{2 \cdot 10^{-7}} \]

6. **Calculating \( P \):**
Substitute the values:
\[ P = \frac{0.4 \times 10^{-4} \cdot \sqrt{3} \cdot 0.008}{2 \cdot 10^{-7}} = \frac{0.4 \cdot 0.008 \cdot \sqrt{3}}{2} \times 10^{3} \]
\[ P = \frac{0.0032 \cdot \sqrt{3}}{2} \times 10^{3} \]
\[ P = 0.0016 \cdot \sqrt{3} \times 10^{3} \]
\[ P = 1.6 \cdot \sqrt{3} \, \text{A m}^2 \]

7. **Final Calculation:**
Using \( \sqrt{3} \approx 1.732 \):
\[ P \approx 1.6 \cdot 1.732 \approx 2.7712 \, \text{A m}^2 \]

### Conclusion:
The value of \( P \) is approximately \( 2.77 \, \text{A m}^2 \).