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Homework 20 Solutions

# Homework 20 Solutions - lim(kl9356 Homework 20...

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lim (kl9356) – Homework 20 – Weathers – (22202) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The magnetic field amplitude of an electro- magnetic wave is 9 . 1 × 10 - 6 T. The speed of light is 2 . 99792 × 10 8 m / s . Calculate the amplitude of the electric field if the wave is traveling in free space. Correct answer: 2728 . 11 N / C. Explanation: Let : c = 2 . 99792 × 10 8 m / s and B = 9 . 1 × 10 - 6 T . According to Maxwell’s equations for an elec- tromagnetic wave in vacuum the relation be- tween the amplitude of the magnetic field and the amplitude of the electric field must be E = c B = (2 . 99792 × 10 8 m / s) (9 . 1 × 10 - 6 T) = 2728 . 11 N / C . The same relation applies even when the ve- locity of light in the medium is not c , in general, E = v B , where v is the velocity of light in that medium. 002 10.0 points Gamma-ray bursters are objects in the uni- verse that emit pulses of gamma rays with high energies. The frequency of the most en- ergetic bursts has been measured at around 3 . 0 × 10 21 Hz. The speed of light is 3 × 10 8 m / s. What is the wavelength of these gamma rays? Correct answer: 1 × 10 - 13 m. Explanation: Let : f = 3 . 0 × 10 21 Hz and c = 3 . 00 × 10 8 m / s . The speed is c = λ = c f = 3 × 10 8 m / s 3 × 10 21 Hz = 1 × 10 - 13 m . 003 (part 1 of 4) 10.0 points A cylindrical filament has radius 2 . 96 mm. Its resistance is 3 . 39 Ω and its length is 0 . 45 m. The filament carries a uniform, steady current of 0 . 829 A. The permeability of free space is 4 π × 10 - 7 N / A 2 . Find the electric field inside the filament. Correct answer: 6 . 24513 N / C. Explanation: Let : = 0 . 45 m , R = 3 . 39 Ω , and I = 0 . 829 A . The electric field inside the filament has to point in the direction of the current; the field causes the current to flow. To find the field, first consider the voltage drop across the ends of the wire, V = I R . The electric field inside the filament is E = V = I R = (3 . 39 Ω) (0 . 829 A) 0 . 45 m = 6 . 24513 N / C . 004 (part 2 of 4) 10.0 points Find B at the surface of the filament. Correct answer: 5 . 60135 × 10 - 5 T. Explanation:

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lim (kl9356) – Homework 20 – Weathers – (22202) 2 Let : r = 2 . 96 mm = 0 . 00296 m , and μ 0 = 4 π × 10 - 7 N / A 2 .
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Homework 20 Solutions - lim(kl9356 Homework 20...

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