Magnetic moment of a short magnet is `16 A - m^(2)` . What is the magnetic induction at a point 40 cm away on its perpendicular bisector line ?

Answer

To find the magnetic induction (B) at a point on the perpendicular bisector of a short magnet, we can use the formula for the magnetic field due to a magnetic dipole. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Magnetic moment (M) = 16 A·m²
- Distance (R) = 40 cm = 0.4 m (convert to meters)

2. **Use the Formula for Magnetic Induction:**
The formula for the magnetic induction (B) at a point on the perpendicular bisector of a magnetic dipole is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{R^3} \]
where:
- \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7}\) T·m/A.

3. **Substitute the Values:**
First, calculate \(\frac{\mu_0}{4\pi}\):
\[ \frac{\mu_0}{4\pi} = 10^{-7} \, \text{T·m/A} \]
Now substitute the values into the formula:
\[ B = 10^{-7} \cdot \frac{2 \cdot 16}{(0.4)^3} \]

4. **Calculate \(R^3\):**
\[ R^3 = (0.4)^3 = 0.064 \, \text{m}^3 \]

5. **Calculate the Magnetic Induction (B):**
Substitute \(R^3\) back into the equation:
\[ B = 10^{-7} \cdot \frac{32}{0.064} \]
Calculate \(\frac{32}{0.064}\):
\[ \frac{32}{0.064} = 500 \]
Now substitute this value:
\[ B = 10^{-7} \cdot 500 = 5 \times 10^{-5} \, \text{T} \]

6. **Final Answer:**
\[ B = 2.5 \times 10^{-7} \, \text{T} \]

### Final Result:
The magnetic induction at a point 40 cm away on the perpendicular bisector line is \(2.5 \times 10^{-7} \, \text{T}\).