College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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**Question:**

6. The electric field amplitude of the signal at that point is 0.780 V/m. What is the amplitude of the magnetic field of the signal at that point?

a) 2.6 nT  
b) 2.1 nT  
c) 1.6 nT  
d) 3.1 nT  
e) 3.6 nT  

**Explanation:**

To find the magnetic field amplitude, we can use the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave, which is given by the equation:

\[ B = \frac{E}{c} \]

where \( c \) is the speed of light in a vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

Given:
- Electric field amplitude (\( E \)) = 0.780 V/m

Therefore:

\[ B = \frac{0.780 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \approx 2.6 \times 10^{-9} \, \text{T} = 2.6 \, \text{nT} \]

Hence, the correct answer is (a) 2.6 nT.
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Transcribed Image Text:**Question:** 6. The electric field amplitude of the signal at that point is 0.780 V/m. What is the amplitude of the magnetic field of the signal at that point? a) 2.6 nT b) 2.1 nT c) 1.6 nT d) 3.1 nT e) 3.6 nT **Explanation:** To find the magnetic field amplitude, we can use the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave, which is given by the equation: \[ B = \frac{E}{c} \] where \( c \) is the speed of light in a vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)). Given: - Electric field amplitude (\( E \)) = 0.780 V/m Therefore: \[ B = \frac{0.780 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \approx 2.6 \times 10^{-9} \, \text{T} = 2.6 \, \text{nT} \] Hence, the correct answer is (a) 2.6 nT.
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