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Question: A compass needle made of pure iron (with density \(8000 Kg m^{-3}\)) has a length \(5 cm\), width \(...

A compass needle made of pure iron (with density 8000Kgm−38000 Kg m^{-3}) has a length 5cm5 cm, width 1mm1 mm and thickness0.5mm0.5 mm. The magnitude of magnetic dipole moment of an iron isμFe=2×10−23JT−1\mu Fe = 2 \times 10^{-23} J T^{-1}, If magnetisation of needle is equivalent to the alignment of 10%10 \% of the atoms in the needle, what is the magnitude of the needle's magnetic dipole moment μ⃗\vec{\mu} ? (mass of iron per mole =0.5Kgmole−10.5 Kg mole^{-1}, NA=6×1023N_{A} = 6 \times 10^{23})
a) 2.4×10−3JT−12.4 \times 10^{-3} J T^{-1}
b) 4.8×10−3JT−14.8 \times 10^{-3} J T^{-1}
c) 7.2×10−3JT−17.2 \times 10^{-3} J T^{-1}
d) 9.6×10−3JT−19.6 \times 10^{-3} J T^{-1}

Explanation

Solution

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=5cm=10−2m5 cm = 10^{-2} m
Width= 1mm=10−3m1 mm = 10^{-3} m
Thickness= 0.5mm=0.5×10−30.5 mm= 0.5 \times 10^{-3}
μFe=2×10−23JT−1\mu Fe = 2 \times 10^{-23} J T^{-1}
ρ=8000Kgm−3\rho = 8000 Kg m^{-3}
Volume = V=5×10−2×10−3×0.5×10−3V = 5 \times 10^{-2} \times 10^{-3} \times 0.5 \times 10^{-3}
Mass of compass needle(m) =volume×densityvolume \times density
m=2.5×10−8×8000=2×10−4m = 2.5 \times 10^{-8} \times 8000 = 2 \times 10^{-4} Kg
m=0.2m=0.2g
now, we have no of atoms in 50g=6×102350 g = 6 \times 10^{23} atoms
no of atoms in 0.2g=0.2×6×1023500.2g = \dfrac{0.2 \times 6 \times 10^{23}}{50}
N = 0.024×10230.024 \times 10^{23}
magnetisation of the needle is equivalent to the alignment of 10%10 \% of the atoms in the needle so we need to multiply the result with 0.10.1 .
μ=μFe×N\mu = \mu Fe \times N
μ=2×10−23×0.024×1023\mu = 2 \times 10^{-23} \times 0.024 \times 10^{23}
μ=4.8×10−3\mu = 4.8 \times 10^{-3}
the magnitude of the needle's magnetic dipole moment μ\mu is μ=4.8×10−3\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%10 \% that magnetisation of needle is equivalent to the alignment of 10%10 \% of the atoms in the needle. Here we need to take 10%10 \% of the total atoms determined by mass.