Problem%20Set%207%20Solutions

# Problem%20Set%207%20Solutions - 49 The saturation...

This preview shows pages 1–5. Sign up to view the full content.

49. The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by M sat = µ n , where n is the number of atoms per unit volume and is the magnetic dipole moment of an atom. The number of nickel atoms per unit volume is n = ρ / m , where is the density of nickel. The mass of a single nickel atom is calculated using m = M / N A , where M is the atomic mass of nickel and N A is Avogadro’s constant. Thus, () ( ) 32 3 22 3 28 3 8.90g cm 6.02 10 atoms mol 9.126 10 atoms cm 58.71g mol 9.126 10 atoms m . A N n M × == The dipole moment of a single atom of nickel is × × M n sat 3 2 Am m 470 10 9126 10 515 10 5 28 24 . . ..

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
N R m = 4 3 3 π ρ . We substitute this into µ total = N to obtain 13 3 total total 3 4 . 34 m R R m ρµ ⎛⎞ π =⇒ = ⎜⎟ π ⎝⎠ The mass of an iron atom is m == × −− 56 56 166 10 9 30 10 27 26 uu k g u k g . bg c h .. Therefore, R = ×× L N M M O Q P P 3 9 30 10 8 0 10 42 1 1 0 18 10 26 22 23 5 . . kg J T kg m J T m. 3 ch c h c hc h π14 10 3 (b) The volume of the sphere is VR s × = × 44 182 10 2 53 10 35 3 16 π 3 π 3 mm 3 and the volume of the Earth is V e = × 4 6 37 10 108 10 6 3 21 π 3 , 3 so the fraction of the Earth’s volume that is occupied by the sphere is 253 10 108 10 23 10 16 21 5 . . × × m m 3 3 53. (a) If the magnetization of the sphere is saturated, the total dipole moment is total = N , where N is the number of iron atoms in the sphere and is the dipole moment of an iron atom. We wish to find the radius of an iron sphere with N iron atoms. The mass of such a sphere is Nm , where m is the mass of an iron atom. It is also given by 4 π R 3 /3, where is the density of iron and R is the radius of the sphere. Thus Nm = 4 π R 3 /3 and
Pr I == × × × × L N M M O Q P P 4 4 22 10 946 10 10 10 4 6 37 10 11 10 24 1 5 2 12 6 2 15 ππ π .. . . . ly m / ly W m W. ch c h c h 14. (a) The power received is () 2 12 22 2 6 m/ 4 1.0 10 W 1.4 10 W. 4 6.37 10 m r P π −− (300 ) = × π× (b) The power of the source would be

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(d) The magnetic field amplitude is 6 8 300V/m 1.0 10 T. 2.998 10 m/s m m E B c == = × × (e) G B must be in the positive z direction when G E is in the positive y direction in order for G G E B × to be in the positive x direction (the direction of propagation). (f) The intensity of the wave is 2 2 22 2 8 0 (300V/m) 119W/m 1.2 10 W/m . 2 2(4 H/m)(2.998 10 m/s) m E I c µπ −7 = × ×10 × (g) Since the sheet is perfectly absorbing, the rate per unit area with which momentum is delivered to it is I / c , so dp dt IA c × () ( . ) .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

### Page1 / 14

Problem%20Set%207%20Solutions - 49 The saturation...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online