A sinusoidal electromagnetic wave of frequency 6.10×1014hz travels in vacuum in the +x direction. the magnetic field is parallel

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3 months ago

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A sinusoidal electromagnetic wave of frequency 6.10×1014hz travels in vacuum in the +x direction. the magnetic field is parallel

to the y axis and has amplitude 5.80×10−4t. part a find the magnitude of the electric field. express your answer to three significant figures with the appropriate units. emax = submitrequest answer part b find the direction of the electric field. find the direction of the electric field. parallel to the + z-axis parallel to the − z-axis parallel to the + y-axis parallel to the + x-axis parallel to the − x-axis parallel to the − y-axis

Physics

1 answer:

Yuliya22 [3.3K] · Answer 1 · 2026-03-21T11:24:06+03:00

Part a) The connection between the electric field and the magnetic field in an electromagnetic wave is
$E=cB$
where
E signifies the strength of the electric field
B indicates the strength of the magnetic field
c represents the speed of light
Using the equation, we determine:
$E=cB=(3 \cdot 10^8 m/s)(5.80 \cdot 10^{-4} T)=1.74 \cdot 10^5 N/C$

Part b) The text does not clarify the orientation of the magnetic field on the y-axis: I speculate it points in the y+ direction.
The direction of the electric field can be established using the right-hand rule, which states:
- the index finger shows the direction of E
- the middle finger indicates the orientation of B
- the thumb reveals the propagation direction of the wave
Because the wave propagates in the x+ direction, and the magnetic field in the y+ direction, we conclude that the electric field direction (index finger) must be z-.