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PolarNik [594]
3 years ago
12

A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i

n the x direction and the magnetic field is in the -y direction.(a) What is the direction of propagation of the wave?(b) If the intensity of the wave at point P is 7.43kW/cm2, what is the electric and magnetic field amplitude at that point? (c = 3.00 ×108 m/s, μ0 = 4π× 10-7 T · m/A,ε0 = 8.85 × 10-12 C2/N · m2)
Physics
1 answer:
olasank [31]3 years ago
7 0

Answer:

a

The direction of the wave propagation is the negative  z -axis

b

The amplitude of  electric and magnetic field are  A_E= 3.35*10^5 V/m ,

A_M= 1.12 *10^{-3} T respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive  x-axis  i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis  (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative  z -axis

             The Intensity of the wave is mathematically represented as

                          I = \frac{1}{2} c \epsilon _O E_{rms}^2

Given that I = 7.43 \frac{kW}{cm^2} =  7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}

Making E_{rms} the subject we have

                   E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }

Substituting values as given on the question

                E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }

                          = 2.37*10^5 \ V/m

The amplitude of the electric field is mathematically represented as

                  A_E = \sqrt{2} * E_{rms}

                         = \sqrt{2} * 2.37*10^5

                        A_E= 3.35*10^5 V/m

The amplitude of the magnetic field is mathematically represented as

                       A_M = \frac{A_E}{c}

Substituting value

                      A_M = \frac{3.35 *10^5}{3.0*10^8}

                             A_M= 1.12 *10^{-3} T

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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
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(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

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The initial kinetic energy is:

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(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

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