The magnetic moment of a short dipole is `1 A m^(2)`.What is the magnitude of the magnetic induction in air at `10 cm` from centre of the dipole on a line making an angle of `30^(@)` from the axis of the dipole?

Explanation

To solve the problem, we need to find the magnitude of the magnetic induction (B) at a distance of 10 cm from the center of a short magnetic dipole with a magnetic moment (M) of 1 A m², at an angle of 30° from the axis of the dipole.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Magnetic moment, \( M = 1 \, \text{A m}^2 \)
- Distance from the dipole, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Angle from the axis, \( \theta = 30^\circ \)

2. **Use the Formula for Magnetic Induction:**
The magnetic induction \( B \) at a distance \( r \) from a dipole at an angle \( \theta \) is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} \cdot \left( \cos^2 \theta + \frac{1}{3} \sin^2 \theta \right) \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (the permeability of free space).

3. **Calculate \( r^3 \):**
\[ r^3 = (0.1)^3 = 0.001 \, \text{m}^3 \]

4. **Calculate \( \cos^2(30^\circ) \) and \( \sin^2(30^\circ) \):**
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
\[ \sin(30^\circ) = \frac{1}{2} \quad \Rightarrow \quad \sin^2(30^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Substitute Values into the Formula:**
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \cdot 1}{0.001} \cdot \left( \frac{3}{4} + \frac{1}{3} \cdot \frac{1}{4} \right) \]
Simplifying, we have:
\[ B = 10^{-7} \cdot \frac{2}{0.001} \cdot \left( \frac{3}{4} + \frac{1}{12} \right) \]

6. **Calculate \( \frac{3}{4} + \frac{1}{12} \):**
To add these fractions, find a common denominator (which is 12):
\[ \frac{3}{4} = \frac{9}{12} \quad \Rightarrow \quad \frac{9}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6} \]

7. **Final Calculation:**
\[ B = 10^{-7} \cdot 2000 \cdot \frac{5}{6} = \frac{10000}{6} \times 10^{-7} = \frac{5000}{3} \times 10^{-7} \approx 1.67 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic induction in air at 10 cm from the center of the dipole on a line making an angle of 30° from the axis of the dipole is approximately \( 1.67 \times 10^{-4} \, \text{T} \).