Question

A sinusoidal electromagnetic wave emitted by a mobile phone has a wavelength of 34.8 cm and an electric-field amplitude of 5.70×10−2 V/m at a distance of 210 m from the phone.
Calculate
(a) the frequency of the wave;
(b) the magnetic-field amplitude;
(c) the intensity of the wave.

Answer 1

Answer:

a) $f = 8.62 \cdot 10^{8} Hz$

b) $B = 1.9 \cdot 10^{-10} T$  

c) $I = 4.30 \cdot 10^{-6} W/m^{2}$

Explanation:

a) The frequency (f) of the wave can be found as follows:

$f = \frac{c}{\lambda}$

Where:

c: is the speed of light = 3x10⁸ m/s

λ: is the wavelength = 34.8 cm

$f = \frac{3 \cdot 10^{8} m/s}{0.348 m} = 8.62 \cdot 10^{8} Hz$

b) The magnetic-flied amplitude (B) is:

$B = \frac{E}{c}$      

Where:

E: is the electric field amplitude = 5.70x10⁻² V/m

$B = \frac{E}{c} = \frac{5.70 \cdot 10^{-2} V/m}{3 \cdot 10^{8} m/s} = 1.9 \cdot 10^{-10} T$  

c) The intensity of the wave (I) is the following:

$I = \frac{E*B}{2\mu_{0}}$

Where:

μ₀: is the permeability of free space =  1.26x10⁻⁶ m*kg/(s²A²)  

$I = \frac{E*B}{2\mu_{0}} = \frac{5.70 \cdot 10^{-2} V/m*1.9 \cdot 10^{-10} T}{2*1.26 \cdot 10^{-6} m*kg/((s^{2}A^{2})} = 4.30 \cdot 10^{-6} W/m^{2}$

I hope it helps you!

Answer 2

The frequency of the wave is $8.62\times 10^8\rm\;Hz$, the magnetic-field amplitude is $1.9\times 10^{-10}\rm\;T$, and the intensity of the wave is $4.298\rm\;W/m^2$.

Given information:

A mobile phone emits electromagnetic radiation.

The wavelength of the wave is $\lambda=34.8$ cm.

The electric-field amplitude is  $5.70\times10^{-2}$ V/m.

Phone is at a distance of 210 m.

The speed of the electromagnetic wave is $c=3\times 10^8$ m/s.

(a)

Now, the frequency of the wave will be calculated as,

$f=\dfrac{c}{\lambda}\\f=\dfrac{3\times 10^8}{0.348}\\f=8.62\times 10^8\rm\;Hz$

(b)

The magnetic-field amplitude can be calculated as,

$B=\dfrac{E}{c}\\B=\dfrac{5.70\times10^{-2}}{3\times 10^8}\\B=1.9\times 10^{-10}\rm\;T$

(c)

$\mu_0$ is the permeability of the vacuum. $\mu_0=1.26\times10^{-6} \rm\;\frac{kg-m}{(A^2s^2)}$

The intensity of the wave can be calculated as,

$I=\dfrac{BE}{2\mu_0}\\I=\dfrac{1.9\times10^{-10 }\times5.7\times10^{-2}}{2\times1.26\times10^{-6}}\\I=4.298\rm\;W/m^2$

Therefore, the frequency of the wave is $8.62\times 10^8\rm\;Hz$, the magnetic-field amplitude is $1.9\times 10^{-10}\rm\;T$, and the intensity of the wave is $4.298\rm\;W/m^2$.

For more details, refer to the link:

https://brainly.com/question/1393179