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deff fn [24]
3 years ago
7

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:
tensa zangetsu [6.8K]3 years ago
7 0
Part a) The relationship between the electric field and the magnetic field of an electromagnetic wave is
E=cB
where
E is the intensity of the electric field
B is the intensity of the magnetic field
c is the speed of light
Using the formula, we find:
E=cB=(3 \cdot 10^8 m/s)(5.80 \cdot 10^{-4} T)=1.74 \cdot 10^5 N/C

part b) it is not specified in the text what is the verse of the magnetic field on the y-axis: I assume its direction is y+.
We can find the direction of the electric field by using the right-hand rule, where:
- the index finger gives the direction of E
- the middle finger gives the direction of B
- the thumb gives the direction of propagation of the wave
since the direction of propagation is x+, and the direction of the magnetic field is y+, we can see that the direction of the electric field (index finger) is z-.
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