Question

A compass needle made of pure iron (with density $8000kg/m^3$) has a length 5 cm, width 1.0 mm and thickness $0.50$ mm. The magnitude of magnetic dipole moment of an iron is $\mu Fe=2\times {10}^{-23}$ JiT, If magnetisation of needle is equivalent to the alignment of 10% of the atoms in the needle, what is the magnitude of the needle's magnetic dipole moment $\overrightarrow{\mu }$? (mass of iron per mole $=0.05kg/mole,N_A=6x{10}^{23}$)

• A. $2.4\times {10}^{-3}J/T$
• B. $4.8\times {10}^{-3}J/T$
• C. $7.2\times {10}^{-3}J/T$
• D. $9.6\times {10}^{-3}J/T$

Answer 1

Answer: (B)

$4.8\times {10}^{-3}J/T$

Given: $\rho =8000kg/m^3$

$volume,V=5\ast {10}^{-2}\ast {10}^{-3}\ast 0.5\ast {10}^{-3}{m}^3=2.5\ast {10}^{-8}{m}^3$

$\mu Fe=2\ast {10}^{-23}J/T$

$1mole=50gm=6\ast {10}^{23}atoms$

To find: needle's magnetic dipole moment, $\mu =?$

Solution: mass of compass needle,$M=\rho \ast V$

$M=8000\ast 2.5\ast {10}^{-8}$

$M=2\ast {10}^{-4}Kg$

$M=0.2gm$

now, we have

no of atoms in $50gm=6\ast {10}^{23}atoms$

$==>$ no of atoms in $0.2gm=\frac{0.2\ast 6\ast {10}^{23}}{50}atoms$

$==>$ $N=0.024\ast {10}^{23}atoms$

now, we know that

$\mu =\mu Fe\ast N$

$==>$ $\mu =2\ast {10}^{-23}\ast 0.024\ast {10}^{23}$

$\mu =4.8\ast {10}^{-3}J/T$