Q.Three identical bar magnets are rivetted together at their common centre and lie in the same plane, their axes equally inclined at 60 degrees to one another so that the six half-arms point radially outward at 60-degree intervals (like a six-spoked star). The system is placed at rest in a slowly (spatially) varying magnetic field, and it is observed that the system shows no motion at all. One of the three magnets is oriented vertically, with its North pole at the top arm and its South pole at the bottom arm. Of the remaining two magnets, one lies along the diagonal joining the upper-left and lower-right arms, and the other lies along the diagonal joining the upper-right and lower-left arms, each slanted arm making 60 degrees with the vertical. Determine which pole (North or South) must lie at each of the four slanted arm-ends belonging to these two magnets, so that the system stays at rest.

Concept: why the net moment must vanish

A magnetic dipole of moment $\vec{m}$ in a field $\vec{B}$ feels a torque $\vec{\tau}=\vec{m}\times\vec{B}$ and, in a non-uniform field, a net force $\vec{F}=\nabla(\vec{m}\cdot\vec{B})$. For the whole rivetted system the resultant torque is $\vec{\tau}=\vec{M}\times\vec{B}$ and the resultant force is governed by $\vec{M}\cdot\vec{B}$, where $\vec{M}=\vec{m}_1+\vec{m}_2+\vec{m}_3$ is the net moment. The system is observed to show no motion at all, and this must hold whatever the direction of the slowly-varying field. The only way both $\vec{\tau}$ and $\vec{F}$ can vanish for every orientation of $\vec{B}$ is

Why-this-condition fixes the poles

Each magnet is identical, so $|\vec{m}_1|=|\vec{m}_2|=|\vec{m}_3|=m$. Three vectors of equal magnitude add to zero only when they are mutually inclined at $120^\circ$.

Take the plane as the $x\!-\!y$ plane and measure angles anticlockwise from the $+x$-axis. The three magnet axes are 60 degrees apart:

• Vertical magnet: axis along $90^\circ$ (up) and $270^\circ$ (down).
• Upper-right / lower-left diagonal: axis along $30^\circ$ and $210^\circ$.
• Upper-left / lower-right diagonal: axis along $150^\circ$ and $330^\circ$.

Steps

1. The vertical magnet has North at the top, so its moment points up: $\vec{m}_1$ is along $90^\circ$, i.e. $\vec{m}_1=m(0,\,1)$.
2. For $\vec{M}=0$ the other two moments must lie at $90^\circ+120^\circ=210^\circ$ and $90^\circ-120^\circ=-30^\circ\equiv330^\circ$.
3. Direction $210^\circ$ is an allowed end of the upper-right/lower-left magnet (it points toward the lower-left arm). So $\vec{m}_2$ points to the lower-left: North pole at the lower-left arm, South pole at the upper-right arm. …

Method: Finding Unknown Pole Orientations from a "Rest in Any Field" (Zero Net Moment) Condition

Use this whenever a rigid assembly of magnets is stated to feel no torque and no net force in a slowly-varying field of arbitrary direction, and you must work out how the individual magnets must be oriented.

Steps

Step 1: Translate "no motion in an arbitrary field" into a condition on the total moment

A magnetic dipole $\vec m$ in a field $\vec B$ feels a torque $\vec\tau=\vec m\times\vec B$ and, in a non-uniform field, a force governed by $\vec F=\nabla(\vec m\cdot\vec B)$. For a rigid group of magnets, replace $\vec m$ with the vector sum $\vec M=\sum_i\vec m_i$. The system shows no motion no matter which way $\vec B$ points at that instant, and both $\vec\tau$ and $\vec F$ must vanish for every possible direction of $\vec B$ — the only vector for which that is true for all $\vec B$ is the zero vector itself:

Step 2: Use the equal-magnitude symmetry to find the required angular spacing

If all the individual magnets are identical, $|\vec m_i|=m$ for every $i$. A set of equal-magnitude vectors sums to zero only for specific symmetric arrangements — e.g. two equal vectors cancel only if antiparallel (180° apart); three equal vectors cancel only if mutually 120° apart; and so on. Identify which symmetric arrangement applies to the number of magnets given.

Step 3: Anchor the arrangement using the one direction you're told …