The average dipole moment of Fe atoms is 1.8 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. Calculate the volume of the iron rod: $V = \pi r^2 L = \pi (0.5 \ cm)^2 	imes 10 \ cm = 2.5 \pi \ cm^3 \approx 7.85 \ cm^3$.$2$. Calculate t

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) $1$. Calculate the volume of the iron rod: $V = \pi r^2 L = \pi (0.5 \ cm)^2 	imes 10 \ cm = 2.5 \pi \ cm^3 \approx 7.85 \ cm^3$.$2$. Calculate the mass of the rod: $m = 	ext{density} 	imes V = 7.87 \ g/cm^3 	imes 7.85 \ cm^3 \approx 61.8 \ g$.$3$. Calculate the number of atoms $(N)$: $N = \frac{m}{M} 	imes N_A = \frac{61.8}{55.87} 	imes 6.023 	imes 10^{23} \approx 6.66 	imes 10^{23}$ atoms.$4$. Calculate the total magnetic moment: $M_{total} = N 	imes \mu_{atom} = (6.66 	imes 10^{23}) 	imes (1.8 	imes 10^{-23} \ A-m^2) \approx 12 \ A-m^2$.

The average dipole moment of $Fe$ atoms is $1.8 \times 10^{-23} \ A-m^2$. The magnetic moment of an iron rod of length $10 \ cm$ and diameter $1 \ cm$ is........$A-m^2$: (density and atomic weight of $Fe$ are $7.87 \ g/cm^3$ and $55.87$ respectively).

Similar Questions

Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is

$A$ magnetic needle is kept in a non-uniform magnetic field. It experiences

Define the pole strength of a magnet and write its $SI$ unit.

The torque on a bar magnet due to the earth's magnetic field is maximum when the axis of the magnet is

The following figures show the arrangement of bar magnets in different configurations. Each magnet has a magnetic dipole moment $\vec{m}$. Which configuration has the highest net magnetic dipole moment?

Vedclass Test Series

Exam Paper Generator

Online Exam Module