The dipole moment associated with an atom of iron in an iron...

The dipole moment associated with an atom of iron in an iron bar has magnitude $2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T}$. Assume that all the atoms in the bar, which is $5.0 \mathrm{~cm}$ long and has a cross-sectional area of $1.0 \mathrm{~cm}^{2}$, have their dipole moments aligned. (a) What is the magnitude of the dipole moment of the bar? (b) What is the magnitude of the torque that must be exerted to hold this magnet perpendicular to an external field of $1.5 \mathrm{~T}$ ? (The density of iron is $7.9 \mathrm{~g} / \mathrm{cm}^{3} .$ )

Key Concepts

Volume, Density, and Atomic Calculations

This concept involves using geometric properties such as volume and material density to calculate the mass and subsequently the number of constituent atoms in a macroscopic object. It is essential for bridging microscopic properties (per-atom measurements) to the overall macroscopic behavior of a material.

Alignment of Atomic Dipoles

Alignment of atomic dipoles is the process by which individual magnetic moments of atoms in a material are oriented in the same direction, leading to a cumulative effect that produces a net magnetic moment. This concept is fundamental in understanding how ferromagnetic materials achieve significant magnetization.

Magnetic Dipole Moment

A magnetic dipole moment is a measure of the magnetic strength and orientation of a magnetic object or particle. It can describe both atomic-scale magnets and larger magnetic systems, and it quantifies how an object interacts with an external magnetic field.

Torque on a Magnetic Dipole

Torque on a magnetic dipole refers to the rotational force that a dipole experiences when placed in an external magnetic field. This torque tends to align the dipole moment with the field, and its magnitude is given by the product of the dipole moment, the field strength, and the sine of the angle between them.

Transcript

00:01 High in the given problem, a dipole moment of an atom of iron that is given as 2 .1 into 10 dash bar minus 23 joules per tesla.

00:28 Length of the bar, iron bar, that is given as 5 .0 centimeter and its area of cross -section.

00:52 That is given as a is equal to 1 .0 centimeter cube so its volume will be given by v is equal to area of cross -section multiplied by the length means it will come out to be 5 .0 into 1 means 5 .0 centimeter cube now in the first part of the problem density of iron is given as d is equal to 7 .9 gram per centimeter cube so mass of this bar will be given by m is equal to volume into density so here it will be volume is 5 centimeter cube and density is 7 .9 gram per centimeter cube so canceling this centimeter cube we get the mass of this bar, iron bar, to be equal to 39 .5 gram.

02:06 Now the molar mass of the iron is 56 gram per mole.

02:19 So the number of moles here, number of moles of iron in this given bar is these are given as mass upon molar mass so it will come out to be for mass this is 39 .5 gram and for molar mass this is 56 gram per mole hence the number of moles here will come out to be 0 .705 moles so number of atoms of iron total atoms will be given by number of moles 7 0 .705 multiplied by our gadrose number so here it will be 0 .705 into 6 .023 into 10 dash to par 23.

03:31 So finally here it will come out to be 4 .25 into 10 dash per 23.

03:40 So dipole moment of this bar will be iron bar will be given by the product of total atoms 4 .25 into 10 dash per 23 with the dipole moment of one one atom, single atom, which was given as 2 .1 into 10 -power minus 23 joules per tesla.

04:15 So finally this magnetic moment in saturation, dipole moment of iron bar in saturation will be given by 8 .9 to 5 joules per tesla which is the answer for the first part of of this problem...

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