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 solve the problem of finding the magnetic moment of a short bar magnet given the magnetic induction field strength at a distance on the equatorial line, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula**:
The magnetic field strength \( B \) at the equatorial point of a short bar magnet is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{m}{R^3} \]
where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( m \) is the magnetic moment of the magnet,
- \( R \) is the distance from the magnet.

2. **Substituting Known Values**:
From the problem, we know:
- \( B = 20 \times 10^{-6} \, \text{T} \)
- \( R = 0.20 \, \text{m} \)

We can substitute these values into the formula.

3. **Rearranging the Formula**:
To find the magnetic moment \( m \), we rearrange the formula:
\[ m = B \cdot \frac{4\pi}{\mu_0} \cdot R^3 \]

4. **Calculating \( R^3 \)**:
First, we calculate \( R^3 \):
\[ R^3 = (0.20)^3 = 0.008 \, \text{m}^3 \]

5. **Substituting \( \mu_0 \)**:
We know \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \). Thus:
\[ \frac{4\pi}{\mu_0} = \frac{4\pi}{4\pi \times 10^{-7}} = \frac{1}{10^{-7}} = 10^7 \]

6. **Putting It All Together**:
Now substituting \( B \), \( R^3 \), and \( \frac{4\pi}{\mu_0} \) into the equation for \( m \):
\[ m = (20 \times 10^{-6}) \cdot (10^7) \cdot (0.008) \]

7. **Calculating the Result**:
\[ m = 20 \times 10^{-6} \times 10^7 \times 0.008 \]
\[ = 20 \times 0.008 \times 10^1 \]
\[ = 0.16 \times 10^1 = 1.6 \, \text{A m}^2 \]

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