Question:
An electromagnetic wave travels from vacuum into a non-magnetic dielectric medium with permittivity $\epsilon = 4\epsilon_0$. If the ratio ($E_0 / B_0$) of the electric field amplitude $E_0$ to the magnetic field amplitude $B_0$ in vacuum is equal to the speed of light $c$, then the corresponding ratio in the given medium is
An electromagnetic wave travels from vacuum into a non-magnetic dielectric medium with permittivity $\epsilon = 4\epsilon_0$. If the ratio ($E_0 / B_0$) of the electric field amplitude $E_0$ to the magnetic field amplitude $B_0$ in vacuum is equal to the speed of light $c$, then the corresponding ratio in the given medium is
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The refractive index of a non-magnetic medium is $n = \sqrt{\epsilon_r} = \sqrt{\frac{\epsilon}{\epsilon_0}}$.
The speed of light in the medium is $v = \frac{c}{n}$.
Since $\epsilon_r = 4$, we get $n = 2$, which immediately yields $v = \frac{c}{2}$.
The speed of light in the medium is $v = \frac{c}{n}$.
Since $\epsilon_r = 4$, we get $n = 2$, which immediately yields $v = \frac{c}{2}$.
Updated On: Jun 16, 2026
- $c / 2$
- $c$
- $c / \sqrt{2}$
- $c / 4$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem asks for the ratio of the electric field amplitude to the magnetic field amplitude of an electromagnetic wave inside a specific dielectric medium.
This ratio is fundamentally linked to the speed of wave propagation in that medium.
Step 2: Key Formula or Approach:
For any electromagnetic wave propagating in a medium, the ratio of the electric field amplitude to the magnetic field amplitude is equal to the wave speed in that medium:
\[ \frac{E}{B} = v \]
The speed of light in a medium is given by:
\[ v = \frac{1}{\sqrt{\mu \epsilon}} \]
Step 3: Detailed Explanation:
• In vacuum, the permeability is $\mu_0$ and the permittivity is $\epsilon_0$. The speed of the wave is:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \frac{E_0}{B_0} \]
• For the non-magnetic dielectric medium:
- Non-magnetic means the magnetic permeability is equal to that of vacuum, i.e., $\mu = \mu_0$.
- The permittivity is given as $\epsilon = 4\epsilon_0$.
• Let us calculate the speed of the wave in this medium:
\[ v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 (4\epsilon_0)}} \]
\[ v = \frac{1}{2\sqrt{\mu_0 \epsilon_0}} \]
• Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, we can write:
\[ v = \frac{c}{2} \]
• Therefore, the ratio of the amplitudes in this medium is:
\[ \frac{E_{\text{medium}}}{B_{\text{medium}}} = v = \frac{c}{2} \]
Step 4: Final Answer:
The corresponding ratio in the given medium is $c / 2$.
The problem asks for the ratio of the electric field amplitude to the magnetic field amplitude of an electromagnetic wave inside a specific dielectric medium.
This ratio is fundamentally linked to the speed of wave propagation in that medium.
Step 2: Key Formula or Approach:
For any electromagnetic wave propagating in a medium, the ratio of the electric field amplitude to the magnetic field amplitude is equal to the wave speed in that medium:
\[ \frac{E}{B} = v \]
The speed of light in a medium is given by:
\[ v = \frac{1}{\sqrt{\mu \epsilon}} \]
Step 3: Detailed Explanation:
• In vacuum, the permeability is $\mu_0$ and the permittivity is $\epsilon_0$. The speed of the wave is:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \frac{E_0}{B_0} \]
• For the non-magnetic dielectric medium:
- Non-magnetic means the magnetic permeability is equal to that of vacuum, i.e., $\mu = \mu_0$.
- The permittivity is given as $\epsilon = 4\epsilon_0$.
• Let us calculate the speed of the wave in this medium:
\[ v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 (4\epsilon_0)}} \]
\[ v = \frac{1}{2\sqrt{\mu_0 \epsilon_0}} \]
• Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, we can write:
\[ v = \frac{c}{2} \]
• Therefore, the ratio of the amplitudes in this medium is:
\[ \frac{E_{\text{medium}}}{B_{\text{medium}}} = v = \frac{c}{2} \]
Step 4: Final Answer:
The corresponding ratio in the given medium is $c / 2$.
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