The pole strength of a horse shoe magnet is `90Am` and distance between the poles is `6 cm`. The magnetic induction at mid point of the line joining the poles is,

Options

A. `10^(-2)T`
B. zero
C. `2xx10^(-2)T`
D. `10^(-4)T`

Explanation

To find the magnetic induction at the midpoint of the line joining the poles of a horseshoe magnet, we can use the formula for the magnetic field due to a magnetic dipole. The magnetic induction (B) at a point along the axial line of a magnetic dipole 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 pole strength,
- \( r \) is the distance from the midpoint to one of the poles.

Given:
- Pole strength \( m = 90 \, Am \)
- Distance between the poles \( d = 6 \, cm = 0.06 \, m \)
- Distance from the midpoint to either pole \( r = \frac{d}{2} = \frac{0.06}{2} = 0.03 \, m \)

### Step-by-Step Solution:

1. **Identify the given values**:
- Pole strength \( m = 90 \, Am \)
- Distance between the poles \( d = 6 \, cm = 0.06 \, m \)
- Midpoint distance \( r = 0.03 \, m \)

2. **Use the formula for magnetic induction**:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} \]

3. **Substitute the values into the formula**:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 90}{(0.03)^3} \]
The \( 4\pi \) cancels out:
\[ B = 10^{-7} \cdot \frac{180}{(0.03)^3} \]

4. **Calculate \( (0.03)^3 \)**:
\[ (0.03)^3 = 0.000027 \, m^3 \]

5. **Substitute this value back into the equation**:
\[ B = 10^{-7} \cdot \frac{180}{0.000027} \]

6. **Calculate \( \frac{180}{0.000027} \)**:
\[ \frac{180}{0.000027} \approx 6666666.67 \]

7. **Now calculate \( B \)**:
\[ B = 10^{-7} \cdot 6666666.67 \approx 0.66666667 \, T \]

8. **Convert to scientific notation**:
\[ B \approx 6.67 \times 10^{-1} \, T \]

### Final Answer:
The magnetic induction at the midpoint of the line joining the poles is approximately \( 0.67 \, T \).