A bar magnet of length `10cm` has a pole strength of `20Am`. Find the magnetic field strength at a distance of `10cm` from its centre on (i) its axial line and on (ii) its equatorial line.

Explanation

To solve the problem, we need to find the magnetic field strength at a distance of 10 cm from the center of a bar magnet on both its axial line and equatorial line. We will use the formulas for the magnetic field strength for each case.

### Given Data:
- Length of the bar magnet, \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Pole strength, \( p = 20 \, \text{Am} \)
- Distance from the center, \( d = 10 \, \text{cm} = 0.1 \, \text{m} \)

### (i) Magnetic Field Strength on the Axial Line

1. **Formula for Magnetic Field Strength on the Axial Line**:
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2p}{d^2 - \left(\frac{l}{2}\right)^2} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

2. **Substituting Values**:
- \( d = 0.1 \, \text{m} \)
- \( l = 0.1 \, \text{m} \) so \( \frac{l}{2} = 0.05 \, \text{m} \)
\[ B_1 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \cdot 20}{(0.1)^2 - (0.05)^2} \]
Simplifying gives:
\[ B_1 = 10^{-7} \cdot \frac{40}{0.01 - 0.0025} = 10^{-7} \cdot \frac{40}{0.0075} \]

3. **Calculating**:
\[ B_1 = 10^{-7} \cdot 5333.33 = 5.33 \times 10^{-4} \, \text{T} \]

### (ii) Magnetic Field Strength on the Equatorial Line

1. **Formula for Magnetic Field Strength on the Equatorial Line**:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{p}{d^2 + \left(\frac{l}{2}\right)^2} \]

2. **Substituting Values**:
\[ B_2 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{20}{(0.1)^2 + (0.05)^2} \]
Simplifying gives:
\[ B_2 = 10^{-7} \cdot \frac{20}{0.01 + 0.0025} = 10^{-7} \cdot \frac{20}{0.0125} \]

3. **Calculating**:
\[ B_2 = 10^{-7} \cdot 1600 = 1.6 \times 10^{-4} \, \text{T} \]

### Final Answers:
- **Magnetic Field Strength on the Axial Line**: \( B_1 = 5.33 \times 10^{-4} \, \text{T} \)
- **Magnetic Field Strength on the Equatorial Line**: \( B_2 = 1.6 \times 10^{-4} \, \text{T} \)