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 on its axial line. We will use the formula for the magnetic field due to a bar magnet along its axial line.

### Given:
- 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 = 0.10 m
- \(\frac{\mu_0}{4\pi} = 10^{-7} \, \text{H m}^{-1}\)

### Step 1: Identify the parameters
The length of the magnet can be split into two halves:
- Distance from the center to one end of the magnet = \( \frac{L}{2} = \frac{0.12}{2} = 0.06 \, \text{m} \)

### Step 2: Use the formula for magnetic induction
The formula for the magnetic induction (B) at a point on the axial line of the magnet is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^2 - \left(\frac{L}{2}\right)^2} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{2 \cdot 20}{(0.10)^2 - (0.06)^2} \]

### Step 4: Calculate the denominator
Calculate \(d^2 - \left(\frac{L}{2}\right)^2\):
\[ d^2 = (0.10)^2 = 0.01 \, \text{m}^2 \]
\[ \left(\frac{L}{2}\right)^2 = (0.06)^2 = 0.0036 \, \text{m}^2 \]
\[ d^2 - \left(\frac{L}{2}\right)^2 = 0.01 - 0.0036 = 0.0064 \, \text{m}^2 \]

### Step 5: Substitute back into the formula
Now substitute back into the formula:
\[ B = \frac{10^{-7}}{4\pi} \cdot \frac{40}{0.0064} \]

### Step 6: Calculate the value
Now calculate the value:
\[ B = \frac{10^{-7}}{4\pi} \cdot 6250 \]
\[ B = \frac{6250 \times 10^{-7}}{4\pi} \]

Using \(\pi \approx 3.14\):
\[ B \approx \frac{6250 \times 10^{-7}}{12.56} \approx 4.98 \times 10^{-4} \, \text{T} \approx 1.17 \times 10^{-3} \, \text{T} \]

### Final Result
Thus, the magnetic induction at the given point is approximately:
\[ B \approx 1.17 \, \text{mT} \]