In the figure, P is a small magnetised needle at the point O. The arr…

The correct answer is Option (4) → PQ₄

The potential energy of the pair is the energy of Q in the field of P, \( U=-\mathbf{m}_Q\cdot\mathbf{B}_P \). It is highest when \( \mathbf{m}_Q \) points against \( \mathbf{B}_P \) and the field is strongest. At the same distance, the field of P on its axis is twice the field on its normal bisector. P points upwards, so Q₄ and Q₆ lie on its axis. Q₃ and Q₅ lie on its normal bisector.

Option (1) PQ₂ — Incorrect. Q₂ is on the normal bisector, where the field of P points downwards. Q₂ points sideways, at \( 90^\circ \) to this field. So \( U=0 \), even though Q₂ is closer to P.

Option (2) PQ₅ — Incorrect. On the normal bisector the field points downwards, and Q₅ points up. So \( U \) is positive here. But this field is only half the axial field, so \( U \) is half the value for PQ₄.

Option (3) PQ₆ — Incorrect. Below P on the axis, the field points upwards, and Q₆ also points up. Q₆ is parallel to the field, so \( U \) is negative. This arrangement has the lowest energy.

Option (4) PQ₄ — Correct. Above P on the axis, the field points upwards, but Q₄ points down. It is anti-parallel to the stronger axial field, so \( U=+\frac{\mu_0}{4\pi}\frac{2m^{2}}{r^{3}} \), the highest value.

NCERT (2026-27):
Example 5.2 – “Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression …:”

    \( \displaystyle \mathbf{B}_P=-\frac{\mu_0}{4\pi}\frac{\mathbf{m}_P}{r^{3}} \quad \text{(on the normal bisector)} \)

    \( \displaystyle \mathbf{B}_P=\frac{\mu_0}{4\pi}\frac{2\mathbf{m}_P}{r^{3}} \quad \text{(on the axis)} \)

“where m_P is the magnetic moment of the dipole P. Equilibrium is stable when m_Q is parallel to B_P, and unstable when it is anti-parallel to B_P.”

FIGURE 5.4