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

4.23 \times 10^{- 8} T

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1.41 \times 10^{- 8} T

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2.83 \times 10^{- 8} T

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0.70 \times 10^{- 8} T

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Solution

The correct option is D $2.83 \times 10^{- 8} T$
$E_{r m s} = 6 V / m$
$E_{r m s} = E_{r m s} \times \sqrt{2} = 6 \sqrt{2} V / m$
Now, $E = C B$
$\to B = \frac{E}{C} = \frac{6 \sqrt{2}}{3 \times 10^{8}}$
$= 2.25 \times 10^{- 8} T$
$= 2.828 \times 10^{- 8} T$

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In an electromagnetic wave in free space the root mean square value of the electric filed is Erms=6V/m. The peak value of the magnetic field is:

The correct option is D 2.83×10−8TErms=6V/mErms=Erms×√2=6√2V/mNow, E=CB→B=EC=6√23×108=2.25×10−8T=2.828×10−8T

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

The correct option is D $2.83 \times 10^{- 8} T$
$E_{r m s} = 6 V / m$
$E_{r m s} = E_{r m s} \times \sqrt{2} = 6 \sqrt{2} V / m$
Now, $E = C B$
$\to B = \frac{E}{C} = \frac{6 \sqrt{2}}{3 \times 10^{8}}$
$= 2.25 \times 10^{- 8} T$
$= 2.828 \times 10^{- 8} T$