A bar magnet with a length of 1 cm and a cross-sectional area of...

The magnetic field at the end-on position of a short bar magnet is $B_{axial} = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}$.

Substituting the given values:

$1.5 \times 10^{-4} = 10^{-7} \times \dfrac{2M}{(0.15)^3}$

$M = \dfrac{1.5 \times 10^{-4} \times 0.003375}{2 \times 10^{-7}} \approx 2.53 \text{ A m}^2$

The volume of the magnet is $V = A \times l = (1.0 \times 10^{-4}) \times 10^{-2} = 10^{-6} \text{ m}^3$.

The magnetization $I$ is:

$I = \dfrac{M}{V} = \dfrac{2.53}{10^{-6}} = 2.53 \times 10^6 \text{ A/m}$

The pole strength $m$ is:

$m = \dfrac{M}{l} = \dfrac{2.53}{10^{-2}} = 253 \text{ A m}$

The magnetic intensity $H$ at the centre due to the two poles (at a distance $l/2 = 0.5 \times 10^{-2} \text{ m}$ each) is directed opposite to the magnetization:

$H = -2 \times \dfrac{1}{4\pi} \dfrac{m}{(l/2)^2} = -\dfrac{2 \times 253}{4\pi(0.005)^2} = -\dfrac{506}{\pi \times 10^{-4}} \approx -1.61 \times 10^6 \text{ A/m}$

The magnetic field $B$ at the centre of the magnet is:

$B = \mu_0(I + H) = 4\pi \times 10^{-7} \times (2.53 \times 10^6 - 1.61 \times 10^6)$

$B = 4\pi \times 10^{-7} \times 0.92 \times 10^6 \approx 1.16 \text{ T} \approx 1.2 \text{ T}$