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Magnetism And Matter

Question
CBSEENPH12037821

Each atom of an iron bar (5 cm x 1 cm x 1 cm) has a magnetic moment 1.8 x 10–23Am2.
(a)    What will be the magnetic moment of the bar in the state of magnetic saturation.
(b)    What will be the torque required to place this magnetised bar perpendicular to magnetic field of 15000 gauss?
Density of iron = 7.8 x 103 kg/m3
Atomic weight of iron = 56
Avagadro’s number = 6.023 x 1023 gm/mole

Solution
Here, volume of specimen, V = 5×1×1=5×10-6m3 

Number of atoms per unit volume 

               n= NA/ρ = ρNA 

For iron,

 A = 56,   ρ = 7.8 × 103 kg/m3N = 6.026 × 1026 kg/mole   

Therefore,
               n = 7.8 × 103 × 6.02 × 102656   = 8.38 × 1028 m-3 

  Total number of atoms in the iron bar
                            N = n V    = 8.38 × 1028 × 5 × 10-6N = 4.19 × 1023 

 Saturated magnetic moment of the bar
                                  M =4.19 × 1023×1.8 × 10-23     = 7.54  Am2 b.)Torque acting on the bar, τ = MB sin θ                                                         = 7.54 × (15000 × 10-4) sin 90°

                                    τ = 11.31 Nm.

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