For a bar magnet, the magnetic field on the axial line at a distance \( r \) from the centre is given by the formula:
\[
B = \frac{{\mu_0 \, m}}{{4 \pi \, r^3}} \left( 2 \right)
\]
where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)),
- \( m \) is the magnetic pole strength,
- \( r \) is the distance from the center of the magnet to the point where the field strength is measure(D)
Given:
- \( B = 10^{-4} \, \text{T} \),
- \( r = 0.1 \, \text{m} \),
- The length of the magnet is 2 cm, so the pole strength \( m \) is to be determine(D)
Substitute the given values into the formula:
\[
10^{-4} = \frac{{(4 \pi \times 10^{-7}) \, m}}{{4 \pi \, (0.1)^3 \times 2}}
\]
Simplifying the equation:
\[
10^{-4} = \frac{{10^{-7} \, m}}{{8 \times 10^{-3}}}
\]
Solving for \( m \):
\[
m = 10^{-4} \times 8 \times 10^{-3} = 25 \, \text{Am}
\]
Thus, the pole strength of the magnet is \( 25 \, \text{Am} \).