Problem 2138
Question
A new system of unit is evolved in which the values of \(\mu_{0}\) and \(\varepsilon_{0}\) are 2 and 8 respectively. Then the speed of light in this system will be (A) \(0.25\) (B) \(0.5\) (C) \(0.75\) (D) 1
Step-by-Step Solution
Verified Answer
The speed of light in the new system of units can be found using the formula \(c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\). Substituting the given values of \(\mu_0 = 2\) and \(\varepsilon_0 = 8\), we get \(c = \frac{1}{\sqrt{(2)(8)}} = \frac{1}{\sqrt{16}} = \frac{1}{4}\). The correct answer is (A) \(0.25\).
1Step 1: Write down the formula for the speed of light in terms of permeability and permittivity constants
In this step, we will write down the formula relating the speed of light to the permeability and permittivity constants: \(c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\).
2Step 2: Substitute the given values
Now we will substitute the given values of \(\mu_0\) and \(\varepsilon_0\) into the formula: \(c = \frac{1}{\sqrt{(2)(8)}}\).
3Step 3: Calculate the speed of light
Let's simplify the expression to find the value of the speed of light: \(c = \frac{1}{\sqrt{16}} = \frac{1}{4}\).
4Step 4: Choose the correct answer from the options
Since the speed of light is \(\frac{1}{4}\), the correct answer is (A) \(0.25\).
Key Concepts
Understanding the Permeability ConstantExploring the Permittivity ConstantUnit System Conversion and Its Importance
Understanding the Permeability Constant
The permeability constant, also known as the magnetic constant, measures how a material can support the formation of a magnetic field. It's essential for understanding how magnetic fields interact with materials.
- Symbol: Often denoted by \( \mu_0 \).
- Role in Physics: In a vacuum, \( \mu_0 \) is a fundamental constant used to calculate magnetic forces in the equations of electromagnetism.
- Use in Speed of Light Formula: The permeability constant is crucial for the speed of light equation: \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \). It interacts with permittivity to determine how fast electromagnetic waves travel through a medium.
Exploring the Permittivity Constant
The permittivity constant, or electric constant, indicates how an electric field affects, and is affected by, a dielectric medium. This concept is vital for understanding electrical interactions.
- Symbol: Represented as \( \varepsilon_0 \).
- Functionality: It defines the ability of a material to permit electric field lines. In free space, it serves as a reference point for measuring permittivity.
- Connection to Speed of Light: In the speed of light formula \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \), \( \varepsilon_0 \) is paired with \( \mu_0 \) to determine light's velocity in a medium.
Unit System Conversion and Its Importance
Unit system conversion isn't just about changing numbers. It involves understanding how these conversions impact the relationships between physical constants and measurements.
- Why It's Necessary: Different systems may use different base units (like meters or feet), affecting calculations. Conversion ensures consistency and accuracy.
- In the Exercise: The problem gives new values for \( \mu_0 \) and \( \varepsilon_0 \), essentially creating a new unit system. Without conversion, this would radically alter our calculations.
- Impact on Formulas: Adjusting unit values impacts directly on formulas such as the speed of light equation, ensuring that the derived values remain useful and understandable.
Other exercises in this chapter
Problem 2136
What is the ratio of velocities of light rays of wavelengths \(4000^{\circ} \mathrm{A}\) and \(8000^{\circ} \mathrm{A}\) in vacuum? (A) \(1: 2\) (B) \(1: 1\) (C
View solution Problem 2137
Which of the following rays are not electromagnetic waves? (A) \(\alpha\) rays (B) \(\gamma\) rays (C) \(\beta\) rays (D) heat rays
View solution Problem 2139
Our eyes respond to wavelength ranging from (A) \(400 \mathrm{~nm}\) to \(700 \mathrm{~nm}\) (B) \(-\infty\) to \(+\infty\) (C) \(1 \mathrm{~mm}\) to \(700 \mat
View solution Problem 2140
In microwave oven, we use electromagnetic oscillators which produce electromagnetic waves in the wavelength range (A) \(1 \mathrm{~mm}\) to \(10 \mathrm{~m}\) (
View solution