College Physics 2e

# Chapter 24

### Problems & Exercises

3.

150 kV/m

6.

(a) 33.3 cm (900 MHz) 11.7 cm (2560 MHz)

(b) The microwave oven with the smaller wavelength would produce smaller hot spots in foods, corresponding to the one with the frequency 2560 MHz.

8.

26.96 MHz

10.

$5.0×10145.0×1014$ Hz

12.
$λ = c f = 3 . 00 × 10 8 m/s 1 . 20 × 10 15 Hz = 2 . 50 × 10 – 7 m λ = c f = 3 . 00 × 10 8 m/s 1 . 20 × 10 15 Hz = 2 . 50 × 10 – 7 m$
14.

0.600 m

16.

(a) $f=cλ=3.00×108 m/s1×10-10 m=3×1018 Hzf=cλ=3.00×108 m/s1×10-10 m=3×1018 Hz$

(b) X-rays

19.

(a) $6.00×106 m6.00×106 m$

(b) $4.33×10−5 T4.33×10−5 T$

21.

(a) 1.50 × 10 6 Hz, AM band
(b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation.

23.

(a) $1.55×1015 Hz1.55×1015 Hz$

(b) The shortest wavelength of visible light is 380 nm, so that

$λ visible λ UV = 380 nm 193 nm = 1.97 . λ visible λ UV = 380 nm 193 nm = 1.97 .$

In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate!

25.

$3 . 90 × 10 8 m 3 . 90 × 10 8 m$

27.

(a) $1.50×1011 m1.50×1011 m$

(b) $0.500 μs0.500 μs$

(c) 66.7 ns

29.

(a) $−3.5×102 W/m2−3.5×102 W/m2$

(b) 88%

(c) $1.7μT1.7μT$

30.
$I = cε 0 E 0 2 2 = 3.00 × 10 8 m/s 8.85 × 10 –12 C 2 /N ⋅ m 2 1 25 V/m 2 2 = 20. 7 W/m 2 I = cε 0 E 0 2 2 = 3.00 × 10 8 m/s 8.85 × 10 –12 C 2 /N ⋅ m 2 1 25 V/m 2 2 = 20. 7 W/m 2$
32.

(a) $I=PA=Pπr2=0.250×10−3 Wπ0.500×10−3 m2=318 W/m2I=PA=Pπr2=0.250×10−3 Wπ0.500×10−3 m2=318 W/m2$

(b) $Iave = cB022μ0⇒B0=2μ0Ic1/2 = 24π×10−7 T⋅m/A318.3 W/m23.00×108 m/s1/2 = 1.63×10−6 T Iave = cB022μ0⇒B0=2μ0Ic1/2 = 24π×10−7 T⋅m/A318.3 W/m23.00×108 m/s1/2 = 1.63×10−6 T$

(c) $E0 = cB0=3.00×108 m/s1.633×10−6 T = 4.90×102 V/m E0 = cB0=3.00×108 m/s1.633×10−6 T = 4.90×102 V/m$

34.

(a) 89.2 cm

(b) 27.4 V/m

36.

(a) 333 T

(b) $1.33×1019W/m21.33×1019W/m2$

(c) 13.3 kJ

38.

(a) $I=PA=P4πr2∝1r2I=PA=P4πr2∝1r2$

(b) $I∝E02, B02⇒E02, B02∝1r2⇒E0,B0∝1rI∝E02, B02⇒E02, B02∝1r2⇒E0,B0∝1r$

40.

13.5 pF

42.

(a) $4.07 kW/m24.07 kW/m2$

(b) 1.75 kV/m

(c) $5.84μT5.84μT$

(d) 2 min 19 s

44.

(a) $5.00×103 W/m25.00×103 W/m2$

(b) $3.88×10−6 N3.88×10−6 N$

(c) $5.18×10−12 N5.18×10−12 N$

46.

(a) $t=0t=0$

(b) $7.50×10−10 s7.50×10−10 s$

(c) $1.00×10−9 s1.00×10−9 s$

48.

(a) $1.01×106 W/m21.01×106 W/m2$

(b) Much too great for an oven.

(c) The assumed magnetic field is unreasonably large.

50.

(a) $2.53×10−20H2.53×10−20H$

(b) L is much too small.

(c) The wavelength is unreasonably small.

53.

(a)

(b) E would increase.

(c) It would be by 1/c.

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