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The electromagnetic wave that delivers a cellular phone callto
a car has a magnetic field with an rms value of 1.5x10^-10 T.The
waves passes perpendicularly through an open window, the areaof
which is 0.20m^2. How much energy does this wave carry throughthe
windown during a 45s phone call? P=energy/time.

Respuesta :

whitneytr12

Answer:

[tex]E=2.41\cdot 10^{-5} J[/tex]

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

[tex]I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}[/tex] (1)

  • c is the speed of light (3*10⁸ m/s)
  • μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
  • B₀ is the magnetic field

[tex]I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}[/tex]

[tex]I_{average}=2.68\cdot 10^{-6} W/m^{2}[/tex]

Now, let's define the relationship between power (P) and average intensity (I).

[tex]I_{average}=\frac{P}{A}[/tex]

  • P is the power
  • A is the area crossed

So we can calculate the power.

[tex]P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W[/tex]

Finally, energy is the product of P times time, so:

[tex]E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J[/tex]

I hope it helps you!

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