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Question

In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is:

The correct answer is

1 : 1

Understanding Energy Density in Electromagnetic Waves

An electromagnetic wave, like light, consists of oscillating electric and magnetic fields that travel through space or a medium. These fields carry energy. The energy carried by the wave is distributed between the electric and magnetic fields.

The energy stored per unit volume in the electric field is called the electric energy density, denoted by \(u_E\). It is given by the formula:

\(u_E = \frac{1}{2} \epsilon_0 E^2\)

where \(\epsilon_0\) is the permittivity of free space and \(E\) is the instantaneous electric field strength.

Similarly, the energy stored per unit volume in the magnetic field is called the magnetic energy density, denoted by \(u_B\). It is given by the formula:

\(u_B = \frac{1}{2 \mu_0} B^2\)

where \(\mu_0\) is the permeability of free space and \(B\) is the instantaneous magnetic field strength.

Ratio of Electric and Magnetic Energy Densities

In an electromagnetic wave propagating in free space, there is a specific relationship between the magnitudes of the electric and magnetic fields. For a plane electromagnetic wave, the instantaneous electric field magnitude \(E\) and magnetic field magnitude \(B\) are related by:

\(E = c B\)

where \(c\) is the speed of light in free space. The speed of light \(c\) is related to \(\epsilon_0\) and \(\mu_0\) by the equation:

\(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

This equation can be rearranged to \(c^2 = \frac{1}{\mu_0 \epsilon_0}\), or \(\mu_0 = \frac{1}{c^2 \epsilon_0}\), or \(\epsilon_0 = \frac{1}{c^2 \mu_0}\).

Now, let's find the ratio of the electric energy density to the magnetic energy density:

\(\frac{u_E}{u_B} = \frac{\frac{1}{2} \epsilon_0 E^2}{\frac{1}{2 \mu_0} B^2}\)

\(\frac{u_E}{u_B} = \frac{\epsilon_0 E^2}{\frac{1}{\mu_0} B^2} = \epsilon_0 \mu_0 \frac{E^2}{B^2}\)

We know that \(E = cB\), so \(E^2 = c^2 B^2\). Substitute this into the ratio expression:

\(\frac{u_E}{u_B} = \epsilon_0 \mu_0 \frac{c^2 B^2}{B^2}\)

The \(B^2\) terms cancel out:

\(\frac{u_E}{u_B} = \epsilon_0 \mu_0 c^2\)

We also know that \(c^2 = \frac{1}{\mu_0 \epsilon_0}\), which means \(\epsilon_0 \mu_0 = \frac{1}{c^2}\). Substitute this into the ratio expression:

\(\frac{u_E}{u_B} = \left(\frac{1}{c^2}\right) c^2\)

\(\frac{u_E}{u_B} = 1\)

So, the ratio of the energy densities of the electric and magnetic fields in an electromagnetic wave is 1:1.

Revision Table: Electromagnetic Wave Properties

Property Electric Field (\(E\)) Magnetic Field (\(B\))
Energy Density Formula \(u_E = \frac{1}{2} \epsilon_0 E^2\) \(u_B = \frac{1}{2 \mu_0} B^2\)
Relationship in EM Wave \(E = cB\) \(B = E/c\)
Ratio \(u_E / u_B\) 1 : 1
Average Energy Density \(\langle u_E \rangle = \frac{1}{4} \epsilon_0 E_0^2\) \(\langle u_B \rangle = \frac{1}{4 \mu_0} B_0^2\)

Additional Information: Energy Flow

The total energy density in an electromagnetic wave is the sum of the electric and magnetic energy densities: \(u = u_E + u_B\). Since \(u_E = u_B\), the total energy density is \(u = 2 u_E = 2 u_B\).

The energy flow in an electromagnetic wave is described by the Poynting vector, \(\mathbf{S}\), which represents the energy transmitted per unit area per unit time (power per unit area). It is given by:

\(\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B})\)

The magnitude of the Poynting vector represents the intensity of the electromagnetic wave. The average intensity is given by \(\langle S \rangle = c \langle u \rangle\), where \(\langle u \rangle\) is the average total energy density.

The fact that the energy densities of the electric and magnetic fields are equal is a fundamental characteristic of electromagnetic waves in vacuum or non-dispersive media.

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The correct answer is

1 : 1

Energy density of electric field: \( u_E = \frac{1}{2} \varepsilon_0 E^2 \) 
Energy density of magnetic field: \( u_B = \frac{1}{2} \frac{B^2}{\mu_0} \)

In free space, \( \frac{u_E}{u_B} = 1 \)

Answer:

1 : 1

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The correct answer is

1 : 1

Energy Density Formulas:

Electric field energy density: \[ u_E = \frac{1}{2}\epsilon_0 E^2 \]

Magnetic field energy density: \[ u_B = \frac{1}{2\mu_0} B^2 \]

For an electromagnetic wave:

\[ E = cB \] and \[ c = \frac{1}{\sqrt{\mu_0\epsilon_0}} \]

Calculate the Ratio:

\[ \frac{u_E}{u_B} = \frac{\frac{1}{2}\epsilon_0 E^2}{\frac{1}{2\mu_0} B^2} = \epsilon_0\mu_0\frac{E^2}{B^2} \]

Substitute E = cB: \[ = \epsilon_0\mu_0\frac{(cB)^2}{B^2} = \epsilon_0\mu_0 c^2 \]

Using \( c^2 = \frac{1}{\mu_0\epsilon_0} \): \[ = \epsilon_0\mu_0 \cdot \frac{1}{\mu_0\epsilon_0} = 1 \]

Final Answer:

The ratio of energy densities is \[ \boxed{1 : 1} \].

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Important Questions from Electromagnetic Waves

  1. What will be the time taken by light to travel 2cm thickness of glass of refractive index 1.5?

  2. Calculate the ratio of the range of a transmitting antenna of height 64 m to the receiving antenna of height 16 m:

  3. A plane electromagnetic wave of frequency 30 MHz travels in free space along the x-direction. At a particular point in space and time, \(\vec{E}\)=6.3j^​V/m. The \(\vec{B}\) at this point would be:

  4. Match List - I with List - II

    List-IList-II
    (A) Microwave(I) Radar System for Aircraft Navigation
    (B) UV Rays(II) To study crystal structure
    (C) X-Rays(III) Radioactive decay of Nucleus
    (D) Gamma-Rays(IV) Lasik eye surgery

    Choose the correct answer from the options given below:

  5. When a capacitor is subjected to a D.C. source it takes a small time interval to get fully charged up. During this small-time interval, there is no passage of charge through the dielectric, yet we use a term - displacement current. This term is used because:

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