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Question: A domain in ferromagnetic iron in the form of cube is having 5 x 1010 atoms. If the side length of t...

A domain in ferromagnetic iron in the form of cube is having 5 x 1010 atoms. If the side length of this domain is 1.5 μ\mu m and each atom has a dipole moment of 8 × 10-24 A m2, then magnetisation of domain is

A

11.8 x 105 Am-1

B

1.18 x 104Am-1

C

11.8 x 104 A m_1

D

1.18 x 10s A m-1

Answer

1.18 x 10s A m-1

Explanation

Solution

: The volume of the cubic domain is,

=3.38×10−12 cm3= 3.38 \times 10 ^ { - 12 } \mathrm {~cm} ^ { 3 }

Number of atoms in domain,

(N)=5×1010( N ) = 5 \times 10 ^ { 10 } atoms

Since each iron atom has a dipole moment (m)

= total number of dipole moment for all atoms

=N×m= \mathrm { N } \times \mathrm { m }

=5×1010×8×10−24= 5 \times 10 ^ { 10 } \times 8 \times 10 ^ { - 24 }

=40×10−14=4×10−13Am2= 40 \times 10 ^ { - 14 } = 4 \times 10 ^ { - 13 } \mathrm { Am } ^ { 2 }

Now the consequent magnetization,

=4×10−13Am23.38×10−18 m3=1.18×105Am−1= \frac { 4 \times 10 ^ { - 13 } \mathrm { Am } ^ { 2 } } { 3.38 \times 10 ^ { - 18 } \mathrm {~m} ^ { 3 } } = 1.18 \times 10 ^ { 5 } \mathrm { Am } ^ { - 1 }