The magnetic induction field strength due to a short bar magnet at a distance 0.20 m on the equatorial line is `20 xx 10^(-6) T`. The magnetic moment of the bar magnet is

Options

A. `3.2 A m^2`
B. `6.4 A m^2`
C. `1.6 A m^2`
D. `16 A m^2`

Answer

To find the magnetic moment \( M \) of the bar magnet given the magnetic induction field strength \( B \) at a distance \( D \) on the equatorial line, we can use the formula for the magnetic field due to a short bar magnet:

\[ B = \frac{\mu_0 M}{4 \pi D^3} \]

Where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( M \) is the magnetic moment,
- \( D \) is the distance from the magnet.

### Step-by-Step Solution:

1. **Identify the given values:**
- \( B = 20 \times 10^{-6} \, \text{T} \)
- \( D = 0.2 \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

2. **Substitute the values into the formula:**
\[ 20 \times 10^{-6} = \frac{4\pi \times 10^{-7} \cdot M}{4 \pi (0.2)^3} \]

3. **Simplify the equation:**
The \( 4\pi \) terms cancel out:
\[ 20 \times 10^{-6} = \frac{10^{-7} \cdot M}{(0.2)^3} \]

4. **Calculate \( (0.2)^3 \):**
\[ (0.2)^3 = 0.008 \, \text{m}^3 \]

5. **Rearrange the equation to solve for \( M \):**
\[ M = (20 \times 10^{-6}) \cdot (0.008) \cdot 10^{7} \]

6. **Calculate \( M \):**
\[ M = 20 \times 0.008 \times 10^{1} = 0.16 \times 10^{1} = 1.6 \, \text{A m}^2 \]

### Final Answer:
The magnetic moment \( M \) of the bar magnet is:
\[ M = 1.6 \, \text{A m}^2 \]