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The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0= 510 nT. Amplitude of the electric field part of the wave is
- a)153 N/C
- b)163N/C
- c)158N/C
- d)173N/C
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The amplitude of the magnetic field part of a harmonic electromagnetic...
Magnetic field part of a harmonic electromagnetic wave in vacuum
,B0=510×10−9T
Speed of light,
C=3×108m/s
E=cBo=153N/C
,B0=510×10−9T
Speed of light,
C=3×108m/s
E=cBo=153N/C
Most Upvoted Answer
The amplitude of the magnetic field part of a harmonic electromagnetic...
Magnetic Field Amplitude:
The given information states that the amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT.
Electric Field Amplitude:
To find the amplitude of the electric field part of the wave, we can use the relationship between the electric field and magnetic field in an electromagnetic wave. The amplitude of the electric field (E0) is related to the amplitude of the magnetic field (B0) by the equation:
E0 = c * B0
Where c is the speed of light in vacuum, which is approximately 3 x 10^8 m/s.
Calculating the Electric Field Amplitude:
Using the given value of B0 = 510 nT and the speed of light in vacuum, we can calculate the amplitude of the electric field:
E0 = (3 x 10^8 m/s) * (510 x 10^-9 T)
E0 = (3 x 510 x 10^-1) V/m
E0 = 153 V/m
Therefore, the amplitude of the electric field part of the wave is 153 V/m.
Conclusion:
So, the correct answer is option 'A' which states that the amplitude of the electric field part of the wave is 153 N/C.
The given information states that the amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT.
Electric Field Amplitude:
To find the amplitude of the electric field part of the wave, we can use the relationship between the electric field and magnetic field in an electromagnetic wave. The amplitude of the electric field (E0) is related to the amplitude of the magnetic field (B0) by the equation:
E0 = c * B0
Where c is the speed of light in vacuum, which is approximately 3 x 10^8 m/s.
Calculating the Electric Field Amplitude:
Using the given value of B0 = 510 nT and the speed of light in vacuum, we can calculate the amplitude of the electric field:
E0 = (3 x 10^8 m/s) * (510 x 10^-9 T)
E0 = (3 x 510 x 10^-1) V/m
E0 = 153 V/m
Therefore, the amplitude of the electric field part of the wave is 153 V/m.
Conclusion:
So, the correct answer is option 'A' which states that the amplitude of the electric field part of the wave is 153 N/C.
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Community Answer
The amplitude of the magnetic field part of a harmonic electromagnetic...
Since E=c*B , the amplitude of electric field will be 3*10^8*51*10^-9
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The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum isB0= 510 nT. Amplitude of the electric field part of the wave isa)153 N/Cb)163N/Cc)158N/Cd)173N/CCorrect answer is option 'A'. Can you explain this answer?
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