(1) The period of oscillation of a magnetic needle in a magnetic field is *T*. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be

(a) 2^{1/4}*T *

(b) 2^{1/2}*T *

(c) 2*T *

(d) *T*√3* *

(e) *T/*2

The period of oscillation (*T*) of the single magnetic needle is given by

*T =* 2π√(I*/*mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’* *is the magnetic dipole moment of the needle and ‘B’* *is flux density of the magnetic field.

When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m^{2} + m^{2}) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period *T*_{1} given by

*T*_{1} = 2π√(2I*/*mB√2) = 2π√(I√2*/*mB) = 2^{1/4}*T*

(2) Three identical magnetic needles each *L* metre long and of dipole moment *m* ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is

(a) *m*

(b) 2*m*

(c) 3*m*

(d) 3*m*/2

(e)** **5*m*/2** **

The distance (AD) between the ends of the compound magnet is 2*L*. Since the pole strength is *m/L*, the dipole moment of the compound magnet is* *(*m/L*)2*L = *2*m*

(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below:

Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by

(1) M*/*2π

(2) M*/*π

(3) M(2 + π)/2π

(4) Mπ/(2 + π)

The above question appeared in EAMCET 2009 (Medicine) question paper.

The distance between the poles of the resulting magnet is (L/2) + 2R

Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π

Therefore, length of the resulting magnet (L/2) + (L/π)

The pole strength (p) of the magnet is given by

p = M/L

Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π

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