Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]} $\hat{i} .$ What is the amplitude of the magnetic field part of the wave?

Solution

electric field , E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]} $\overset{⏞}{i}$ .

We know that
$c = \frac{E_{0}}{B_{0}}$

$∴$ $B_{0} = \frac{E_{0}}{c} = \frac{3.1}{3 \times 10^{8}} = 1.03 \times 10^{- 8} T$

$= 0.0103 \times 10^{- 6} T = 0.0103 μT$
is the required amplitude of magnetic field.

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Current Electricity

Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]} $\hat{i} .$ What is the amplitude of the magnetic field part of the wave?

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Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}i^. What is the amplitude of the magnetic field part of the wave?

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Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s)t]} $\hat{i} .$ What is the amplitude of the magnetic field part of the wave?