931_PartUniversity Physics Solution

# 931_PartUniversity Physics Solution - 32-4 Chapter 32 2 rad...

This preview shows pages 1–3. Sign up to view the full content.

32-4 Chapter 32 (b) 22 r a d 0.0174 rad/m 361 m k ππ λ == = (c) 36 2 (2 )(830 10 Hz) 5.22 10 rad/s f ωπ π × = × (d) Eq.(32.18): 81 1 max max (2.998 10 m/s)(4.82 10 T) 0.0144 V/m Ec B × × = EVALUATE: This wave has a very long wavelength; its frequency is in the AM radio braodcast band. The electric and magnetic fields in the wave are very weak. 32.12. IDENTIFY: max max . B = SET UP: The magnetic field of the earth is about 4 10 T. EXECUTE: 3 11 8 3.85 10 V/m 1.28 10 T. 3.00 10 m/s E B c × = × × EVALUATE: The field is much smaller than the earth’s field. 32.13. IDENTIFY and SET UP: vf = relates frequency and wavelength to the speed of the wave. Use Eq.(32.22) to calculate n and K . EXECUTE: (a) 8 7 14 2.17 10 m/s 3.81 10 m 5.70 10 Hz v f × × (b) 8 7 14 2.998 10 m/s 5.26 10 m c f × × (c) 8 8 1.38 × = × c n v (d) m so (1.38) 1.90 nK K KK n =≈ = EVALUATE: In the material and vc f < is the same, so is less in the material than in air. < always, so n is always greater than unity. 32.14. IDENTIFY: Apply Eq.(32.21). max max E cB = . = . Apply Eq.(32.29) with m0 K μμ = in place of 0 μ . SET UP: 3.64 K = . m 5.18 K = EXECUTE: (a) 8 7 m (3.00 10 m s) 6.91 10 m s. (3.64)(5.18) c v KK × = × (b) 7 6 6.91 10 m s 1.06 10 m. 65.0 Hz v f × (c) 3 10 max max 7 7.20 10 V m 1.04 10 T. 6.91 10 m s E B v × = × × (d) 31 0 82 max max 0 ( 7 . 2 01 0 Vm ) ( 1 . 0 41 0 T ) 5.75 10 W m . ( 5 . 1 8 ) EB I K −− ×× = × EVALUATE: The wave travels slower in this material than in air. 32.15. IDENTIFY: / IPA = . 2 1 0m a x 2 Ic E = P . max max E cB = . SET UP: The surface area of a sphere of radius r is 2 4 Ar = . 12 2 2 0 8.85 10 C /N m = ×⋅ P . EXECUTE: (a) 2 (0.05)(75 W) 330 W/m 4( 3 . 0 m ) P I A = × . (b) 2 max 12 2 2 8 0 ( 3 3 0 W / m ) 500 V/m (8.85 10 C /N m )(3.00 10 m/s) I E c = × P . 6 max max 1.7 10 T 1.7 T E B c × = . EVALUATE: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb. Our calculation approximates the filament as a point source that radiates uniformly in all directions. 32.16. IDENTIFY and SET UP: The direction of propagation is given by × ! ! E B . EXECUTE: (a) && & & () . =×− =− S i j k (b) && & & . =×= S ji k (c) &&& & =− ×− = () . S ki j (d) && & & . =×− = S ik j EVALUATE: In each case the directions of E ! , B ! and the direction of propagation are all mutually perpendicular.

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

View Full Document
Electromagnetic Waves 32-5 32.17. IDENTIFY: max max E cB = . × !! E B is in the direction of propagation. SET UP: 8 3.00 10 m/s c . max 4.00 V/m. E = EXECUTE: 8 max max 1.33 10 T BE c == × . For E ! in the + x -direction, × ! ! E B is in the + z -direction when B ! is in the + y -direction. EVALUATE: E ! , B ! and the direction of propagation are all mutually perpendicular. 32.18. IDENTIFY: The intensity of the electromagnetic wave is given by Eq.(32.29): 22 1 0m a x 0r m s 2 . Ic E c E PP The total energy passing through a window of area A during a time t is IAt . SET UP: 12 0 8.85 10 F/m P EXECUTE: 21 2 8 2 2 0r m s Energy (8.85 10 F m)(3.00 10 m s)(0.0200 V m) (0.500 m )(30.0 s) 15.9 J μ × × = cE At P EVALUATE: The intensity is proportional to the square of the electric field amplitude. 32.19. IDENTIFY and SET UP: Use Eq.(32.29) to calculate I , Eq.(32.18) to calculate max , B and use 2 av /4 IP r π = to calculate av . P (a) EXECUTE: 25 2 1 a x m a x 2 ; 0.090 V/m, so 1.1 10 W/m IE E I = × P (b) 10 max max max max so / 3.0 10 T Ec B c = × (c) 2 3 2 av (4 ) (1.075 10 W/m )(4 )(2.5 10 m) 840 W PIr ππ × ×= (d) EVALUATE: The calculation in part (c) assumes that the transmitter emits uniformly in all directions.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

931_PartUniversity Physics Solution - 32-4 Chapter 32 2 rad...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online