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IIT JEE Advanced 2009 Paper 1

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Question : 20

Total: 60

Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

Column I	Column II
(A) E = 0	(P) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is perpendicular to the plane of the hexagon.
(B) V ≠ 0	(Q) Charges are on a line perpendicular to PQ at equal intervals. M is the midpoint between the two innermost charges.
(C) B = 0	(R) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is perpendicular to the plane of the rings.
(D) μ ≠ 0	(S) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the longer sides. M is at the centre of the rectangle. PQ is parallel to the longer sides.
	(T) Charges are placed on two coplanar, identical insulting rings at equal intervals. M is the midpoint between the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings.

[JEE Adv 2009 P1]

(A)→(P), (R), (S); (B)→(R), (S); (C)→(P), (Q), (T); (D)→(P), (S)
(A)→(P), (R), (S); (B)→(R), (S); (C)→(P), (Q), (T); (D)→(R), (S)
(A)→(P), (Q), (S); (B)→(R), (S); (C)→(P), (Q), (T); (D)→(Q), (S)
(A)→(P), (R), (S); (B)→(R), (S); (C)→(P), (R), (T); (D)→(R), (S)

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