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Thepotemich [5.8K]
3 years ago
6

A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction.

Physics
1 answer:
Lostsunrise [7]3 years ago
4 0

Answer:

a) 1.13 10-8 T.  b) +y direction

Explanation:

a)

For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.

At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:

E(z,t) = Emax cos (kz-ωt-φ) +x

B(z,t) = Bmax cos (kz-ωt-φ) +y

In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:

Bmax = Emax/c

So, solving for Bmax:

Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.

b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.

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Answer:

Power = 2 j/s

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2 years ago
If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d
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3 years ago
Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
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by cosine law we know that

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3 years ago
An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat
irina [24]

Explanation:

The charge on the electron is, q=-1.6\times 10^{-19}\ C

The electric field at a distance r from the electron is :

E=k\dfrac{q}{r^2}

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k is the electrostatic constant, k=\dfrac{1}{4\pi \epsilon_o}

We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.

So, the correct option is " the  electric field is directed toward the electron and has a magnitude of k\dfrac{q}{r^2}. Hence, this is the required solution.

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Answer:

194516 sheets

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4 0
3 years ago
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