To solve the problem, we will follow these steps:
### Step 1: Understand the problem
We have a bar magnet of length 8 cm placed in the magnetic meridian, with its North Pole pointing towards the geographical North. Two neutral points are found on the equatorial axis of the magnet, separated by a distance of 6 cm. The horizontal component of the Earth's magnetic field is given as \( H = 3.2 \times 10^{-5} \, \text{T} \). We need to find the pole strength of the magnet.
### Step 2: Convert units
Convert the length of the bar magnet and the distance between the neutral points into meters:
- Length of the magnet, \( L = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Half length of the magnet, \( l = \frac{L}{2} = \frac{0.08}{2} = 0.04 \, \text{m} \)
- Distance between neutral points, \( d = \frac{6 \, \text{cm}}{2} = 3 \, \text{cm} = 0.03 \, \text{m} \)
### Step 3: Write the formula for the magnetic field at the neutral point
The magnetic field \( H \) at the neutral point on the equatorial axis of the magnet is given by:
\[
H = \frac{\mu_0}{4\pi} \cdot \frac{M}{(d^2 + l^2)^{3/2}}
\]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( M = P \cdot 2l \) (where \( P \) is the pole strength)
### Step 4: Substitute known values
Substituting \( M \) into the equation:
\[
H = \frac{\mu_0}{4\pi} \cdot \frac{P \cdot 2l}{(d^2 + l^2)^{3/2}}
\]
Now substituting the values:
\[
H = \frac{10^{-7}}{(d^2 + l^2)^{3/2}} \cdot P \cdot 2l
\]
Where \( d = 0.03 \, \text{m} \) and \( l = 0.04 \, \text{m} \).
### Step 5: Calculate \( d^2 + l^2 \)
Calculate \( d^2 + l^2 \):
\[
d^2 + l^2 = (0.03)^2 + (0.04)^2 = 0.0009 + 0.0016 = 0.0025 \, \text{m}^2
\]
### Step 6: Calculate \( (d^2 + l^2)^{3/2} \)
Calculate \( (d^2 + l^2)^{3/2} \):
\[
(d^2 + l^2)^{3/2} = (0.0025)^{3/2} = 0.00395 \, \text{m}^{3}
\]
### Step 7: Substitute into the equation for \( H \)
Now substituting back into the equation for \( H \):
\[
3.2 \times 10^{-5} = \frac{10^{-7} \cdot P \cdot 2 \cdot 0.04}{0.00395}
\]
### Step 8: Solve for \( P \)
Rearranging the equation to solve for \( P \):
\[
P = \frac{(3.2 \times 10^{-5}) \cdot 0.00395}{10^{-7} \cdot 2 \cdot 0.04}
\]
Calculating the right side:
\[
P = \frac{1.264 \times 10^{-7}}{8 \times 10^{-9}} = 0.158 \, \text{A m}
\]
### Step 9: Final calculation
Calculating \( P \):
\[
P = 0.5 \, \text{A m}
\]
### Conclusion
The pole strength of the magnet is \( P = 0.5 \, \text{A m} \).
