What is the magnetic induction due to a short bar magnet of moment of `0.5 Am^(2)` at a point along its axis at a distance of 20 cm from the centre of the magnet?
`(mu_(0)/(4pi)=10^(-7)" SI unit")`

Options

A. `6.25xx10^(-6)" Wb/m"^(2)`
B. `25xx10^(-6)" Wb/m"^(2)`
C. `12.5xx10^(-6)" Wb/m"^(2)`
D. `18.5xx10^(-6)" Wb/m"^(2)`

Explanation

To solve the problem of finding the magnetic induction (B) due to a short bar magnet at a point along its axis, we can follow these steps:

### Step 1: Understand the given data
We are given:
- Magnetic moment (M) = 0.5 Am²
- Distance (d) = 20 cm = 0.20 m
- The value of \(\frac{\mu_0}{4\pi} = 10^{-7} \, \text{SI unit}\)

### Step 2: Use the formula for magnetic induction along the axis of a dipole
The formula for the magnetic induction (B) at a point along the axis of a short magnetic dipole is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} \]

### Step 3: Substitute the values into the formula
Now, substituting the known values into the formula:
\[ B = \left(10^{-7}\right) \cdot \frac{2 \cdot 0.5}{(0.20)^3} \]

### Step 4: Calculate \(d^3\)
First, calculate \(d^3\):
\[ d^3 = (0.20)^3 = 0.008 \, \text{m}^3 \]

### Step 5: Substitute \(d^3\) back into the equation
Now substitute \(d^3\) back into the equation:
\[ B = 10^{-7} \cdot \frac{1}{0.008} \]

### Step 6: Calculate the fraction
Calculating the fraction:
\[ \frac{1}{0.008} = 125 \]

### Step 7: Final calculation of B
Now substitute this value back into the equation:
\[ B = 10^{-7} \cdot 125 = 1.25 \times 10^{-5} \, \text{T} \]

### Step 8: Convert to Weber per meter square
Since \(1 \, \text{T} = 1 \, \text{Wb/m}^2\), we can express the final answer as:
\[ B = 1.25 \times 10^{-5} \, \text{Wb/m}^2 \]

### Final Answer
Thus, the magnetic induction due to the short bar magnet at the specified point is:
\[ B = 12.5 \times 10^{-6} \, \text{Wb/m}^2 \]

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