IIT-JEE 2009 · Paper 1, Physics question 59

• P (alternating charges on a hexagon): opposite corners carry opposite charges, so each pair's field points towards its negative charge; the three such fields are 120° apart and cancel: $E=0$. The net charge is zero at equal distances: $V=0$. Rotating, the net charge on the single ring is zero, so $B=0$ and $\mu=0$.
• Q (alternating charges on a line): the innermost pair gives a nonzero field at M. Each charge has an opposite charge at the same distance on the other side: $V=0$, and on rotation each circle carries zero net charge: $B=0$, $\mu=0$.
• R (three − on the inner ring, three + on the outer ring): each ring's field cancels at the centre: $E=0$. $V=3kq\left(\frac1{r_2}-\frac1{r_1}\right)\ne0$. Rotation gives opposite currents in rings of different radii, so neither $B$ nor $\mu$ cancels.
• S (− at the corners, + at the long-side midpoints): symmetric pairs cancel: $E=0$; but $V=kq\left(\frac{2\cdot2}a-\frac{4\cdot2}{\sqrt5a}\right)\ne0$. All charges are at distance $\frac a2$ from PQ, with net charge $-2q$, so rotation gives nonzero $B$ and $\mu$.
• T (+ ring and − ring, mirror images): the field at M is nonzero (both rings push/pull the same way). Each + charge has a − charge at the mirror point, at the same distance from M and from PQ, so $V=0$, and their circular currents cancel exactly: $B=0$, $\mu=0$.

Answer: A → P, R, S; B → R, S; C → P, Q, T; D → R, S