The period (T) of angular oscillations of a magnet suspended freely in a magnetic field of flux density ‘B’ is given by

The period of oscillation is given by T = 2π √(I/ mB) with usual notations.

In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.

The equations for the period in the two cases are therefore

T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and

T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2

The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

**T = 2π √(I/ mB)**where ‘I’ is the moment of inertia of the magnet about the axis of the angular motion and ‘m’ is the magnetic dipole moment. You will find questions based on this relation in your entrance examination question papers. Here is a question which is meant also for checking your grasp of the moment of inertia and the vector property of the magnetic dipole moment:**Two identical magnets are placed one above the other and tied together so that their like poles are in contact. The period of oscillation of this combination (on suspending horizontally using torsionless suspension fibre) in a horizontal magnetic field is 2 s. What will be the period of oscillation if the magnets are placed one above the other such that they are mutually perpendicular and bisect each other?**

(a) 2 s (b) √2 s (c) 2√2 s (d) 2

(a) 2 s (b) √2 s (c) 2√2 s (d) 2

^{¼}s (e) 2×2^{¼}sThe period of oscillation is given by T = 2π √(I/ mB) with usual notations.

In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.

The equations for the period in the two cases are therefore

T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and

T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2

^{¼}s.The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:

**The period of oscillation of a magnet at a place is 4 seconds. When it is remagnetised so that the pole strength becomes 4 times the initial value, the period of oscillation in seconds is**

(a) ½ (b) 1 (c) 2 (d) 4

(a) ½ (b) 1 (c) 2 (d) 4

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

NICE

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