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monitta
3 years ago
8

Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned

between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Physics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

Option (2) is correct.

The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.

Explanation:

An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

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The color of the central maximum is white.

Explanation:

A diffraction grating in optics is generally known as an optical component that contains a periodic structure. It is commonly used to breakdown and diffract light into different beams that move in several directions. For example, if a light ray is allowed to pass through the component (i.e. diffraction grating), the   central maximum will have a white color.

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When an object is translated, its image is the same size as the pre-image.<br> a. True<br> b. False
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20 m/s

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4 years ago
An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Nimfa-mama [501]

Explanation:

It is given that,

Speed of the electron in horizontal region, v=1.9\times 10^7\ m/s

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Vertical acceleration, a_y=\dfrac{F_y}{m}

a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}  

a_y=5.37\times 10^{14}\ m/s^2..........(1)

Let t is the time taken by the electron, such that,

t=\dfrac{x}{v_x}

t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}

t=1.26\times 10^{-9}\ s...........(2)

Let d_y is the vertical distance deflected during this time. It can be calculated using second equation of motion:

d_y=ut+\dfrac{1}{2}a_yt^2

u = 0

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3 years ago
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