Question

The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of

• A. 10⁸ ms^-1
• B. 10^-5 ms^-1
• C. 10⁵ ms^-1
• D. 10^-8 ms^-1

Answer 1

The Correct Option is A

Concept: 

For an electromagnetic wave, the ratio of the magnitudes of electric field (\( E \)) and magnetic field (\( B \)) is equal to the speed of light (\( c \)):

\[ \frac{E}{B} = c \]

Calculation:

The speed of electromagnetic waves (speed of light) in vacuum is approximately:

\[ c = 3 \times 10^{8}\,ms^{-1} \]

Given options are approximate orders of magnitude. The correct closest magnitude from the given choices is:

\[ 10^{8}\,ms^{-1} \]

Final Conclusion:
The ratio of electric field to magnetic field for electromagnetic waves is of the order \( 10^{8}\,ms^{-1} \).

Answer 2

The problem asks for the ratio of the magnitudes of the electric field (\(E\)) to the magnetic field (\(B\)) of an electromagnetic wave.

For an electromagnetic wave propagating in vacuum (or air, approximately), the relationship between the magnitudes of the electric field (\(E\)) and the magnetic field (\(B\)) is given by: \[ E = cB \] where \( c \) is the speed of light in vacuum.

The ratio of the magnitudes of the electric field to the magnetic field is therefore: \[ \frac{E}{B} = c \]

The speed of light in vacuum is a fundamental constant, \( c \approx 3 \times 10^8 \, \text{m/s} \).

Therefore, the ratio \( \frac{E}{B} \) is: \[ \frac{E}{B} \approx 3 \times 10^8 \, \text{m/s} \]

The question asks for the order of magnitude of this ratio. The order of magnitude is \( 10^8 \). The units are meters per second (\(\text{m/s}\) or \(\text{ms}^{-1}\)).

So, the ratio is of the order of \( 10^8 \, \text{ms}^{-1} \).