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Kelvi

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In an electromagnetic wave in free space, the root mean square value of the electric field is $E_{rms} = 6$ V/m. The peak value of the magnetic field is

medium

Electromagnetic Waves

2017

physics

$2.83 \times 10^{-8}$ T

$0.70 \times 10^{-8}$ T

$4.23 \times 10^{-8}$ T

$1.41 \times 10^{-8}$ T

Explanation

To solve this problem, we need to find the peak value of the magnetic field $B_0$ in an electromagnetic wave. $\\$ Given: $\\ E_{rms} = 6$

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V/m

\\

The relationship between the root mean square (rms) value and the peak value of the electric field is:

\\ E_{rms} = \frac{E_0}{\sqrt{2}} \\

where

E_0

is the peak value of the electric field.

\\

Rearranging the equation to solve for

E_0: \\ E_0 = E_{rms} \cdot \sqrt{2} \\

Substitute the given value:

\\ E_0 = 6 \cdot \sqrt{2}

V/m

\\ E_0 = 6 \cdot 1.414

V/m

\\ E_0 \approx 8.484

V/m

\\

In free space, the relationship between the electric field

E_0

and the magnetic field

B_0

is given by:

\\ E_0 = c \cdot B_0 \\

where

c

is the speed of light in vacuum, approximately

3 \times 10^8

m/s.

\\

Rearrange to solve for

B_0: \\ B_0 = \frac{E_0}{c} \\

Substitute the known values:

\\ B_0 = \frac{8.484}{3 \times 10^8}

T

\\ B_0 \approx 2.828 \times 10^{-8}

T

\\

Therefore, the peak value of the magnetic field is approximately

2.83 \times 10^{-8}

T.

\\

This corresponds to Option 1.