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3 years ago

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Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structu

re, (2) an atomic radius of 0.154 nm, and (3) a saturation flux density of 0.83 tesla. Determine the number of Bohr magnetons per atom for this material. A: What is the volume, in m3, for this unit cell?B: How many atoms are there per m3 in this material?C: What is the number of Bohr magentons per atom for this material?

1 answer:

scoray [572]3 years ago

3 0

Answer:

a. 2.9218x10^(-29) m^3

b. 7.1230x10^(28) atoms/m^3

c. 2.0812 BM/atom

Explanation:

Atomic radius r = 0.154 nm

saturation flux density Bs = 0.83 tesla

- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3

= (2 x 0.154x10^-9)^3

= 2.9218x10^-29 m^3

- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;

Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)

= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)

= 2.4251x10^-29 / 1.1652x10^-29

= 2.0812 BM/atom

- Number of atoms per Vc, N = nb / Vc

= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms

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Answer:

the angle of twist of B with respect to D is -1.15°

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The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

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1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

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where ;

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shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

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Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

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$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

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$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

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