Physics Solution Manual for 1100 and 2101

In this problem each of the polarizers therefore

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Unformatted text preview: .85 × 10−12 C2 / N ⋅ m 2 ) ( ) c. The rms value Erms of the electric field is related to the peak value E0 by Erms = E0 / 2 . The peak electric field is, therefore, = E0 = 2 Erms 2 (13.4 N/C ) 19.0 N/C = ______________________________________________________________________________ 29. REASONING AND SOLUTION Since the sun emits radiation uniformly in all directions, at a distance r from the sun's center, the energy spreads out over a sphere of surface area 2 4π r . Therefore, according to S = P /(4π r 2 ) (Equation 16.9), the total power radiated by the sun is P = π r 2 ) = W/m 2 )(4π )(1.50 ×1011 m) 2 = ×1026 W S (4 (1390 3.93 ______________________________________________________________________________ 30. REASONING When a stationary charge is placed in an electric field, it experiences an electric force. The magnitude F of the electric force is given by Equation 18.2 as F = q E, where q is the magnitude of the charge and E is the magnitude of the electric field. When a stationary charge is placed in a magnetic field, it does not experience a magnetic force, because the charge is not moving. According to Equation 21.1, the magnitude of the magnetic force is related to the magnitude B of the magnetic field by F = q vB sin θ , where v is the speed of the charge and θ is the angle between the velocity of the charge and the magnetic field. Since the charge is stationary, v = 0 m/s and the magnetic force is zero. When a moving charge is placed in an electric field, it experiences an electric force that is given by Equation 18.2. It does not matter whether the charge is stationary or moving. When a charge moves ( v ≠ 0 m/s ) and its velocity is perpendicular to the magnetic field (θ = 90°), it experiences a magnetic force, as specified by Equation 21.1. Chapter 24 Problems 1293 SOLUTION a. The magnitude of the electric force is F = q E , where the magnitude of the electric field 2 is related to the intensity S of the laser beam by S = cε 0 E (Equation 24.5b). Therefore, the magnitude of the electric force is S cε 0 = qE q F= 2.5 × 10 W/m 2.6 × 10 C 2.5 × 10−5 N = = 8 −12 2 2 3.00 × 10 m/s 8.85 × 10 C / N ⋅ m ( −8 ) 3 ( 2 ( ) ) b. Since the particle is not moving, the magnetic force on it is zero, F = 0 N . c. The electric force on the particle is the same whether it is moving or not, so the answer is 5 2.5 × 10− N . the same as in part (a);= F d. The magnitude of the magnetic force is given by Equation 21.1 as F = q vB sin θ . The magnitude B of the magnetic field is related to the intensity S of the laser beam by S = cB2/µ0 (Equation 24.5c). Thus, the magnetic force is = q vB sin θ q v F= ( −8 )( µ0 S c sin θ = 2.6 × 10 C 3.7 × 10 m/s 4 ) ( 4π × 10 −7 T ⋅ m/A )( 2.5 × 10 W/m 3 3.00 × 10 m/s 8 2 ) sin 90.0° = 3.1 × 10− N 9 ______________________________________________________________________________ 31. REASONING The electromagnetic solar power that strikes an area A⊥ oriented perpendicular to the direction in which the sunlight is radiated is P = SA⊥ , where S is the intensity of the sunlight. In the problem, the solar panels are not oriented perpendicular to the direction of the sunlight, because it strikes the panels at an angle θ with respect to the = normal. We wish to find the solar power that impinges on the solar panels when θ 25° , = given that the incident power is 2600 W when θ 65° . 1294 ELECTROMAGNETIC WAVES SOLUTION When the angle that the sunlight makes with the normal to the solar panel is θ , the power that strikes the solar panel is given by P = SA cos θ , where the area perpendicular to the sunlight is A⊥ = A cosθ (see the drawing). Therefore we can write P2 SA cos θ 2 = P SA cos θ1 1 Normal θ Sunlight Solar panel Area = A A⊥ = A cos θ where the intensity S of the sunlight that reaches the panel, as well as the area A, are the same in both cases. Therefore, we have P2 P1 = cos θ 2 cos θ 1 Solving for P2 , we find that when θ 2 = 35° , the solar power impinging on the panel is cos θ 2 cos 25° = P P2 = (2600 W)= 5600 W 1 cos 65° cos θ1 ______________________________________________________________________________ 32. REASONING AND SOLUTION The intensity S of a wave is the power passing perpendicularly through a surface divided by the area A of the surface. But power is the total energy U per unit time t, so the intensity can be written as S= Total energy U = Time ⋅ Area tA Equation 24.5c relates the intensity S of the electromagnetic wave to the magnitude B of its magnetic field; namely S = (c / µ0 ) B 2 . Combining these two results, we have U c2 B = tA µ0 If the rms value for the magnetic field is used, the energy becomes the average energy U . Thus, the average energy that this wave carries through the window in a 45 s phone call is 3.0 ×108 m/s c2 U = –7 BrmstA =(1.5 ×10 –10 T) 2 (45 s)(0.20 m 2 ) = 4.8 ×10 –5 J 4π ×10 T ⋅ m/A µ0 ______________________________________________________________________________ Chapter 24 Problems 1295 33. REASONING The fraction of the sun’s power that is intercepted by Mercury is...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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