A bar magnet of length `10cm` has a pole strength of `10Am`. Calculate the magnetic field at a distance of `0*2m` from its centre at a point on its (i) axial line (ii) equatorial line.

Explanation

To solve the problem of calculating the magnetic field at a distance of 0.2 m from the center of a bar magnet on its axial and equatorial lines, we will follow these steps:

### Given Data:
- Length of the bar magnet, \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Pole strength, \( m = 10 \, \text{Am} \)
- Distance from the center of the magnet, \( d = 0.2 \, \text{m} \)

### Step 1: Calculate the Magnetic Moment
The magnetic moment \( M \) of the bar magnet can be calculated using the formula:
\[ M = m \times l \]
Substituting the values:
\[ M = 10 \, \text{Am} \times 0.1 \, \text{m} = 1 \, \text{Am}^2 \]

### Step 2: Calculate the Magnetic Field on the Axial Line
The formula for the magnetic field \( B_{\text{axial}} \) at a distance \( d \) from the center of the magnet on its axial line is given by:
\[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^2 - l^2} \]
Where \( \mu_0 = 10^{-7} \, \text{T m/A} \).

Substituting the values:
\[ B_{\text{axial}} = \frac{10^{-7}}{4\pi} \cdot \frac{2 \times 1}{(0.2)^2 - (0.1)^2} \]
Calculating \( (0.2)^2 - (0.1)^2 \):
\[ (0.2)^2 = 0.04, \quad (0.1)^2 = 0.01 \quad \Rightarrow \quad 0.04 - 0.01 = 0.03 \]
Now substituting back:
\[ B_{\text{axial}} = \frac{10^{-7}}{4\pi} \cdot \frac{2}{0.03} \]
Calculating \( \frac{2}{0.03} \):
\[ \frac{2}{0.03} = \frac{200}{3} \approx 66.67 \]
Now substituting this value:
\[ B_{\text{axial}} = \frac{10^{-7} \times 66.67}{4\pi} \]
Calculating \( 4\pi \approx 12.57 \):
\[ B_{\text{axial}} \approx \frac{10^{-7} \times 66.67}{12.57} \approx 5.30 \times 10^{-6} \, \text{T} \]

### Step 3: Calculate the Magnetic Field on the Equatorial Line
The formula for the magnetic field \( B_{\text{equatorial}} \) at a distance \( d \) from the center of the magnet on its equatorial line is given by:
\[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{(d^2 + l^2)^{3/2}} \]
Substituting the values:
\[ B_{\text{equatorial}} = \frac{10^{-7}}{4\pi} \cdot \frac{1}{(0.2^2 + 0.1^2)^{3/2}} \]
Calculating \( 0.2^2 + 0.1^2 \):
\[ 0.2^2 + 0.1^2 = 0.04 + 0.01 = 0.05 \]
Now substituting back:
\[ B_{\text{equatorial}} = \frac{10^{-7}}{4\pi} \cdot \frac{1}{(0.05)^{3/2}} \]
Calculating \( (0.05)^{3/2} \):
\[ (0.05)^{3/2} = \sqrt{0.05^3} = \sqrt{0.000125} \approx 0.01118 \]
Now substituting this value:
\[ B_{\text{equatorial}} = \frac{10^{-7}}{4\pi} \cdot \frac{1}{0.01118} \]
Calculating:
\[ B_{\text{equatorial}} \approx \frac{10^{-7}}{12.57 \times 0.01118} \approx \frac{10^{-7}}{0.140} \approx 7.14 \times 10^{-6} \, \text{T} \]

### Final Results:
- The magnetic field on the axial line is approximately \( 5.30 \times 10^{-6} \, \text{T} \).
- The magnetic field on the equatorial line is approximately \( 7.14 \times 10^{-6} \, \text{T} \).