What is the magnetic induction due to a short bar magnet of magnetic moment `0.5 Am^(2)` at a point along the equator at a distance of 20 cm from the centre of the magnet?

Options

A. `12.5xx10^(-6)" Wb/m"^(2)`
B. `6.25xx10^(-6)" Wb/m"^(2)`
C. `3.5xx10^(-6)" Wb/m"^(2)`
D. `18.5xx10^(-6)" Wb/m"^(2)`

Explanation

To find the magnetic induction (B) due to a short bar magnet at a point along the equator, we can use the formula for the magnetic field at a point along the equatorial line of a magnetic dipole. The formula is given by:

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} \]

Where:
- \( B \) is the magnetic induction,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \)),
- \( M \) is the magnetic moment of the magnet,
- \( r \) is the distance from the center of the magnet to the point where we are measuring the magnetic induction.

### Step 1: Identify the given values
- Magnetic moment, \( M = 0.5 \, Am^2 \)
- Distance, \( r = 20 \, cm = 0.2 \, m \)

### Step 2: Substitute the values into the formula
We substitute \( M \) and \( r \) into the formula for \( B \):

\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \cdot 0.5}{(0.2)^3} \]

### Step 3: Simplify the equation
The \( 4\pi \) in the numerator and denominator cancels out:

\[ B = 10^{-7} \cdot \frac{1}{(0.2)^3} \]

### Step 4: Calculate \( (0.2)^3 \)
Calculate \( (0.2)^3 \):

\[ (0.2)^3 = 0.008 \]

### Step 5: Substitute back to find \( B \)
Now substitute \( 0.008 \) back into the equation:

\[ B = 10^{-7} \cdot \frac{1}{0.008} \]

### Step 6: Calculate \( \frac{1}{0.008} \)
Calculate \( \frac{1}{0.008} \):

\[ \frac{1}{0.008} = 125 \]

### Step 7: Final calculation of \( B \)
Now substitute this value back into the equation for \( B \):

\[ B = 10^{-7} \cdot 125 = 1.25 \times 10^{-5} \, T \]

### Final Answer
The magnetic induction due to the short bar magnet at a point along the equator at a distance of 20 cm from the center of the magnet is:

\[ B = 1.25 \times 10^{-5} \, T \]