The dipole moment of a short bar magnet is `1.25 A-m^(2)`. The magnetic field on its axis at a distance of 0.5 metre from the centre of the magnet is

Options

A. `1.0xx10^(-4)"Newton"//"amp-meter"`
B. `4xx10^(-2)"Newton"//"amp-meter"`
C. `2xx10^(-6)"Newton"//"amp-meter"`
D. `6.64xx10^(-8)"Newton"//"amp-meter"`

Explanation

To find the magnetic field on the axis of a short bar magnet at a distance from its center, we can use the formula for the magnetic field \( B \) on the axis of a magnetic dipole:

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

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( p \) is the magnetic dipole moment,
- \( r \) is the distance from the center of the dipole.

Given:
- \( p = 1.25 \, \text{A m}^2 \)
- \( r = 0.5 \, \text{m} \)

Now, let's substitute the values into the formula.

### Step 1: Substitute the values into the formula

\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 1.25}{(0.5)^3} \]

### Step 2: Simplify the expression

The \( 4\pi \) terms cancel out:

\[ B = 10^{-7} \cdot \frac{2 \times 1.25}{(0.5)^3} \]

### Step 3: Calculate \( (0.5)^3 \)

\[ (0.5)^3 = 0.125 \]

### Step 4: Substitute back into the equation

\[ B = 10^{-7} \cdot \frac{2 \times 1.25}{0.125} \]

### Step 5: Calculate \( \frac{2 \times 1.25}{0.125} \)

\[ \frac{2.5}{0.125} = 20 \]

### Step 6: Final calculation of \( B \)

\[ B = 10^{-7} \cdot 20 = 2 \times 10^{-6} \, \text{T} \]

### Final Answer

The magnetic field on the axis at a distance of 0.5 meters from the center of the magnet is:

\[ B = 2 \times 10^{-6} \, \text{T} \]

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