A small bar magnet has a magnetic moment of m = 2.4 A m²…

(a) The neutral point. With the north pole pointing north, the field of the magnet at the end of the line at right angles points south, against the Earth's field. They cancel where $\dfrac{\mu_0 m}{4\pi d^3} = B_H$. So $d^3 = \dfrac{\mu_0 m}{4\pi B_H} = 0.0080\ \text{m}^3$, and $d = 0.200\ \text{m}$.

(b) The point P. On the line of the magnet the field is $\dfrac{2\mu_0 m}{4\pi d^3}$, which is $60\ \mu\text{T}$ at $d = 0.20\ \text{m}$, and it points north, along the moment of the magnet. The Earth's field also points north, so the two add: $60\ \mu\text{T} + 30\ \mu\text{T} = 90\ \mu\text{T}$.

(c) After the turn. Now the field of the magnet points south on its own line, against the Earth's field, so the neutral points move onto that line. They are where $\dfrac{2\mu_0 m}{4\pi d^3} = B_H$: $d^3 = \dfrac{2\mu_0 m}{4\pi B_H} = 0.0160\ \text{m}^3$, so $d = 0.252\ \text{m}$.

(d) Torque. $\tau = m B_H \sin\theta = 2.4\ \text{A m}^2 \times 30 \times 10^{-6}\ \text{T} \times \sin 30^\circ = 36.0\ \mu\text{N m}$.

(e) Work. The potential energy $-\,m B_H \cos\theta$ goes from $-\,m B_H$ to $+\,m B_H$, so $W = 2 m B_H = 2 \times 2.4\ \text{A m}^2 \times 30 \times 10^{-6}\ \text{T} = 144\ \mu\text{J}$.