Answer:
16.042336 Am²
27.2719712 Nm
Explanation:
Dipole moment association with iron atom
[tex]\frac{\mu}{N}=2.9\times 10^{-23}\ J/T[/tex]
L = Length of bar = 6.5 cm
A = Area of bar = 1 cm²
[tex]N_A[/tex] = Avogadro constant = [tex]6.022\times 10^{23}[/tex]
[tex]\rho[/tex] = 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
[tex]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[/tex]
Dipole moment is given by
[tex]\frac{\mu}{N}\times n\\ =2.9\times 10^{-23}\times 5.53184\times 10^{23}\\ =16.042336\ Am^2[/tex]
The dipole moment of the bar is 16.042336 Am²
Torque is given by
[tex]\tau=\mu Bsin\theta\\\Rightarrow \tau=16.042336\times 1.7sin 90\\\Rightarrow \tau=27.2719712\ Nm[/tex]
The torque exerted to hold this magnet perpendicular to an external field is 27.2719712 Nm
The dipole moment of the bar and the torque required to hold this magnet perpendicular to an external field of 1.7 T is mathematically given as
M=16.042336 Am^2
t=27.2719712 Nm
Question Parameter(s):
dipole moment associated with an iron atom of an iron bar is 2.9 × 10-23 J/T.
the atoms in the bar, which is 6.5 cm long
has a cross-sectional area of 1.0 cm2,
Generally, the equation for the Number of atoms is mathematically given as
[tex]N=\frac{\rho LA}{M}\times N_A[/tex]
Therefore
[tex]N=\frac{7.9* 6.5* 1}{55.9}* 6.022* 10^{23}[/tex]
N=5.53184* 10^{23} atoms
Dipole moment
m=u/N*n
Hence
M=2.9* 10^{-23}* 5.53184* 10^{23}
M=16.042336 Am^2
In conclusion, Torque
t=Bsin\theta
t=16.042336* 1.7sin 90
t=27.2719712 Nm
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