The pole strength of `12 cm` long bar magnet is `20 A m`. The magnetic induction at a point 10 cm away from the centre of the magnet on its axial line is `[(mu_(0))/(4pi)=10^(-7)"H m"^(-1)]`

Options

A. `1.17xx10^(-3)` T
B. `2.20xx10^(-3)` T
C. `1.17xx10^(-2)` T
D. `2.20xx10^(-2)` T

Explanation

To solve the problem, we need to calculate the magnetic induction (B) at a point 10 cm away from the center of a bar magnet that is 12 cm long, with a pole strength of 20 A m. We will use the formula for the magnetic field along the axial line of a bar magnet.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the bar magnet (L) = 12 cm = 0.12 m
- Pole strength (m) = 20 A m
- Distance from the center of the magnet to the point (d) = 10 cm = 0.1 m
- Value of \(\frac{\mu_0}{4\pi} = 10^{-7} \, \text{H m}^{-1}\)

2. **Determine the Position of the Point:**
- The point P is located 10 cm away from the center of the magnet along its axial line.

3. **Use the Formula for Magnetic Field (B) on the Axial Line:**
- The formula for the magnetic field at a distance d from the center of a bar magnet is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^2 - \left(\frac{L}{2}\right)^2} \]
- Here, \(\frac{L}{2}\) is the distance from the center of the magnet to either pole.

4. **Calculate the Values:**
- First, calculate \(\left(\frac{L}{2}\right)^2\):
\[ \frac{L}{2} = \frac{0.12}{2} = 0.06 \, \text{m} \]
\[ \left(\frac{L}{2}\right)^2 = (0.06)^2 = 0.0036 \, \text{m}^2 \]
- Now, substitute the values into the formula:
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{2 \cdot 20}{(0.1)^2 - 0.0036} \]
- Calculate \((0.1)^2\):
\[ (0.1)^2 = 0.01 \, \text{m}^2 \]
- Now substitute this into the equation:
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{40}{0.01 - 0.0036} \]
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{40}{0.0064} \]

5. **Calculate the Final Value of B:**
- Calculate \(\frac{40}{0.0064}\):
\[ \frac{40}{0.0064} = 6250 \]
- Now substitute this back into the equation:
\[ B = \frac{10^{-7}}{4\pi} \cdot 6250 \]
- Using \(\pi \approx 3.14\):
\[ B \approx \frac{10^{-7} \cdot 6250}{4 \cdot 3.14} \approx \frac{6250 \times 10^{-7}}{12.56} \approx 4.98 \times 10^{-4} \, \text{T} \]
- Finally, we find:
\[ B \approx 1.17 \times 10^{-3} \, \text{T} \]

### Final Answer:
The magnetic induction at the point 10 cm away from the center of the magnet on its axial line is approximately \(1.17 \times 10^{-3} \, \text{T}\).