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a nonsinusoidal electromagnetic wave like that described in section 32.2 has uniform electric and magnetic fields. the magnitude of the poynting vector for this wave is 11.0 w/m2w/m2 . find the magnitude of the electric field. express your answer with the appropriate units. ee

Respuesta :

oladayoademosu

The magnitude of the electric field of the electromagnetic wave is E = 64.4 V/m

What is an electromagnetic wave?

An electromagnetic wave is awave that does not require a medium to propagate

What is an electric field?

An electric field is a region in which the effect of an electric force is felt.

How to find the magnitude of the electric field?

Since a nonsinusoidal electromagnetic wave like that described in section 32.2 has uniform electric and magnetic fields. the magnitude of the poynting vector for this wave is 11.0 W/m².

What is the poynting vector?

The magnitude of the poynting vector is given by

S = E²/μ₀c where

  • E = electric field,
  • μ₀ = permeability of free space = 4π × 10⁻⁷ H/m and
  • c = speed of electromagnetic waves = 3 × 10⁸ m/s

Since we require the electric field, making E subject of the formula, we have

E = √(Sμ₀c )

Given that S = 11.0 W/m²,

Substituting the value sof the variables into the equation, we have

E = √(Sμ₀c )

E = √(11.0 W/m² × 4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)

E = √(44π × 3 × 10WH/s)

E = √(1320π WH/s)

E = √(4146.9 WH/s)

E = 64.4 V/m

So, the magnitude of the electric field is E = 64.4 V/m

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