Question

A compass needle made of pure iron (with density $8000 Kg m^{-3}$) has a length $5 cm$, width $1 mm$ and thickness$0.5 mm$. The magnitude of magnetic dipole moment of an iron is$\mu Fe = 2 \times 10^{-23} J T^{-1}$, 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 $\vec{\mu}$ ? (mass of iron per mole =$0.5 Kg mole^{-1}$, $N_{A} = 6 \times 10^{23}$)
a) $2.4 \times 10^{-3} J T^{-1}$
b) $4.8 \times 10^{-3} J T^{-1}$
c) $7.2 \times 10^{-3} J T^{-1}$
d) $9.6 \times 10^{-3} J T^{-1}$

Answer 1

Hint: First find out the mass of the needle for finding the number of atoms so that we can measure how many electrons contribute to the magnetic moment. Using density and dimensions, we can find mass. After that no. of atoms can determine from the given data and take product of magnetic dipole moment with No. of atoms.

Complete answer:
Given information -
Length=$5 cm = 10^{-2} m$
Width= $1 mm = 10^{-3} m$
Thickness= $0.5 mm= 0.5 \times 10^{-3}$
$\mu Fe = 2 \times 10^{-23} J T^{-1}$
$\rho = 8000 Kg m^{-3}$
Volume = $V = 5 \times 10^{-2} \times 10^{-3} \times 0.5 \times 10^{-3}$
Mass of compass needle(m) =$volume \times density$
$m = 2.5 \times 10^{-8} \times 8000 = 2 \times 10^{-4}$ Kg
$m=0.2$g
now, we have no of atoms in $50 g = 6 \times 10^{23}$ atoms
no of atoms in $0.2g = \dfrac{0.2 \times 6 \times 10^{23}}{50}$
N = $0.024 \times 10^{23} $
magnetisation of the needle is equivalent to the alignment of $10 \%$ of the atoms in the needle so we need to multiply the result with $0.1$ .
$\mu = \mu Fe \times N$
$\mu = 2 \times 10^{-23} \times 0.024 \times 10^{23} $
$\mu = 4.8 \times 10^{-3}$
the magnitude of the needle's magnetic dipole moment $\mu$ is $\mu = 4.8 \times 10^{-3}$.

Option (b) is correct

Additional Information:
A magnetic moment is a term that describes the magnetic intensity and magnet orientation or any other object that gives a magnetic field. More clearly, a magnetic moment introduces a magnetic dipole moment, the magnetic moment component that a magnetic dipole can represent, and a magnetic dipole consisting of magnetic north and south pole separated by a minute distance.

Note:
No. of atoms plays an important role, so read the question carefully, there is mention about $10 \%$ that magnetisation of needle is equivalent to the alignment of $10 \%$ of the atoms in the needle. Here we need to take $10 \%$ of the total atoms determined by mass.