Question

If the relative permeability and dielectric constant of a given medium are equal to \(\mu_{\mathrm{r}}\) and \(\mathrm{K}\) respectively, then the refractive index of the medium is equal to (A) \(\sqrt{\left(\mu_{\mathrm{T}} \mathrm{K}\right)}\) (B) \(\sqrt{\left(\mu_{1} E_{0}\right)}\)

Short answer

The correct expression for the refractive index based on the relative permeability \(\mu_r\) and dielectric constant \(K\) of the medium is \[n = \sqrt{(\mu_r \mu_0)(K \epsilon_0)}.\] None of the given options match this expression.

Step-by-step solution

Recall the relationships between permeability and permittivity

To find the refractive index, we need to work with the permeability and permittivity of the medium. Recall that: 1. Relative permeability (\(\mu_r\)) is the ratio of the material's permeability (\(\mu\)) to the permeability of free space (\(\mu_0\)): \[\mu = \mu_r \mu_0\] 2. Dielectric constant (\(K\)) is the ratio of the material's permittivity (\(\epsilon\)) to the permittivity of free space (\(\epsilon_0\)): \[\epsilon = K \epsilon_0\]

Use the formula for the speed of light in a medium

The speed of light in a medium (\(v\)) is given by the equation: \[v = \frac{1}{\sqrt{\mu \epsilon}}\]

Use the formula for the refractive index

The refractive index (\(n\)) of a medium is defined as the ratio of the speed of light in vacuum (\(c\)) to the speed of light in the medium (\(v\)): \[n = \frac{c}{v}\]

Substitute the values for relative permeability and dielectric constant into the formula for the speed of light

We have the relative permeability and the dielectric constant in terms of \(\mu_r\) and \(K\). We can now substitute these values into the speed of light equation: \[v = \frac{1}{\sqrt{(\mu_r \mu_0)(K \epsilon_0)}}\]

Substitute the expression for the speed of light in the refractive index formula

Now substitute the expression for \(v\) from Step 4 into the refractive index formula: \[n = \frac{c}{\frac{1}{\sqrt{(\mu_r \mu_0)(K \epsilon_0)}}}\]

Simplify the expression for the refractive index

After simplifying, the expression for the refractive index becomes: \[n = \sqrt{(\mu_r \mu_0)(K \epsilon_0)}\] Comparing this expression to the given options in the exercise, we see that it does not match any of the options directly. Are any of the given options equivalent to this expression?

Check the given options

Let's analyze the given options: (A) \(\sqrt{(\mu_{\mathrm{T}} \mathrm{K})}\): This option is missing the permittivity (\(\epsilon_0\)) and permeability (\(\mu_0\)) of free space. (B) \(\sqrt{(\mu_{1} E_{0})}\): This option involves an electric field (\(E_0\)) which is not relevant in this situation, and it is also missing the dielectric constant (\(K\)). Given the provided options, none of them correctly represent the refractive index based on the relative permeability and dielectric constant.

Key concepts

Relative Permeability

Relative permeability is a measure of how easily a material can become magnetized. It represents the material's ability to support the formation of a magnetic field within itself. This is a key concept in understanding how materials respond to magnetic fields.

• Relative permeability (\( \mu_r \)) is the ratio of a material's permeability (\( \mu \)) to the permeability of free space (\( \mu_0 \)).
• The formula is: \( \mu = \mu_r \mu_0 \).
• A value of \( \mu_r \) greater than 1 indicates that the material is magnetically conductive, while a value less than 1 suggests that it is less susceptible to become magnetized.

Understanding relative permeability helps clarify the behavior of different materials in magnetic fields and is crucial for applications involving electromagnetism.

Dielectric Constant

The dielectric constant, also known as relative permittivity, is a property that indicates how well a material can store electrical energy in an electric field. It is essential for understanding how materials interact with electric fields.

• The dielectric constant (\( K \)) is the ratio of the material's permittivity (\( \epsilon \)) to the permittivity of free space (\( \epsilon_0 \)).
• The formula is: \( \epsilon = K \epsilon_0 \).
• Materials with a high dielectric constant can store more electrical energy compared to those with a low dielectric constant.

This concept plays a significant role in the design of capacitors and other electrical components, as it affects the efficiency and effectiveness of their operation.

Speed of Light in Medium

The speed of light in a medium is different from the speed of light in a vacuum. This adjustment is due to the medium's optical properties, which are affected by its permeability and permittivity.

• The speed of light (\( v \)) in a medium is given by the formula: \( v = \frac{1}{\sqrt{\mu \epsilon}} \).
• Where \( \mu \) represents the material's permeability and \( \epsilon \) its permittivity.
• Slower speeds occur in more optically "dense" media with higher permittivity and permeability.

The concept of speed of light in various media is critical for understanding how light behaves as it travels through different materials, influencing fields like optics and telecommunications.

Permeability of Free Space

The permeability of free space, also known as the magnetic constant, is a physical constant that expresses how magnetic fields interact with space.

• It is denoted by \( \mu_0 \) and has a value of approximately \( 4\pi \times 10^{-7} \) H/m (henry per meter).
• The permeability of free space is fundamental to Maxwell's equations, which describe how electric and magnetic fields propagate.
• This constant allows scientists and engineers to calculate and predict the behavior of magnetic fields in a vacuum.

Understanding \( \mu_0 \) is essential for studying electromagnetic theory, as it establishes the basis for how magnetic interactions occur in a vacuum.

Permittivity of Free Space

Permittivity of free space, often called the electric constant, is another fundamental physical constant that characterizes how electric fields interact with empty space.

• Represented by \( \epsilon_0 \), with a value of about \( 8.854 \times 10^{-12} \, F/m \) (farads per meter).
• It plays a crucial role in Coulomb’s law and in the equations describing the behavior of electric fields in free space.
• Permitivity of free space helps us understand how much resistance is encountered when forming an electric field in a vacuum.

This concept is foundational in physics and engineering, providing insights into how electric fields are established and how they propagate through free space.