Question

Suppose that the dipole moment associated with an iron atom of an iron bar is 2.9 × 10-23 J/T. Assume that all the atoms in the bar, which is 6.5 cm long and has a cross-sectional area of 1.0 cm2, have their dipole moments aligned. (a) What is the dipole moment of the bar? (b) What torque must be exerted to hold this magnet perpendicular to an external field of 1.7 T? (The density of iron is 7.9 g/cm3 and its molar mass is 55.9 g/mol.)

Answer 1

Answer:

16.042336 Am²

27.2719712 Nm

Explanation:

Dipole moment association with iron atom

$\frac{\mu}{N}=2.9\times 10^{-23}\ J/T$

L = Length of bar = 6.5 cm

A = Area of bar = 1 cm²

$N_A$ = Avogadro constant = $6.022\times 10^{23}$

$\rho$ = Density of iron = 7.9 g/cm³

M = Molar mass = 55.9 g/mol

B = Magnetic field = 1.7 T

Number of atoms is given by

$N=\frac{\rho LA}{M}\times N_A\\\Rightarrow N=\frac{7.9\times 6.5\times 1}{55.9}\times 6.022\times 10^{23}\\\Rightarrow N=5.53184\times 10^{23}\ atoms$

Dipole moment is given by

$\frac{\mu}{N}\times n\\ =2.9\times 10^{-23}\times 5.53184\times 10^{23}\\ =16.042336\ Am^2$

The dipole moment of the bar is 16.042336 Am²

Torque is given by

$\tau=\mu Bsin\theta\\\Rightarrow \tau=16.042336\times 1.7sin 90\\\Rightarrow \tau=27.2719712\ Nm$

The torque exerted to hold this magnet perpendicular to an external field is 27.2719712 Nm