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:
2.1×10−8\(\hat{k}\)T
Understanding Electromagnetic Waves in Free Space
This question asks us to find the magnetic field (\(\vec{B}\)) component of a plane electromagnetic wave in free space, given its frequency, direction of propagation, and the electric field (\(\vec{E}\)) vector at a specific point and time.
Electromagnetic waves consist of oscillating electric (\(\vec{E}\)) and magnetic (\(\vec{B}\)) fields that are perpendicular to each other and also perpendicular to the direction of wave propagation. In free space, these waves travel at the speed of light, \(c\).
Given Information:
- Frequency of the wave, \(f = 30 \text{ MHz} = 30 \times 10^6 \text{ Hz}\). (Note: Frequency is not directly needed for the magnitude calculation, but confirms it's an EM wave).
- Direction of propagation: along the positive x-direction (\(\hat{i}\)).
- Electric field at a point and time, \(\vec{E} = 6.3 \hat{j} \text{ V/m}\). This means the electric field is in the positive y-direction (\(\hat{j}\)).
- The medium is free space.
Key Principles for Electromagnetic Waves
In free space, the magnitudes of the electric and magnetic fields in an electromagnetic wave are related by the speed of light (\(c\)):
\(E = cB\)
where \(E\) is the magnitude of the electric field, \(B\) is the magnitude of the magnetic field, and \(c\) is the speed of light in free space, approximately \(3 \times 10^8 \text{ m/s}\).
The direction of propagation of the electromagnetic wave (\(\hat{v}\)) is given by the cross product of the electric and magnetic field directions:
\(\hat{v} = \hat{E} \times \hat{B}\)
Calculating the Magnetic Field (\(\vec{B}\))
Step 1: Calculate the Magnitude of the Magnetic Field
We are given the magnitude of the electric field, \(E = 6.3 \text{ V/m}\). Using the relation \(E = cB\), we can find the magnitude of the magnetic field \(B\):
\(B = \frac{E}{c}\)
Substitute the values: \(E = 6.3 \text{ V/m}\) and \(c = 3 \times 10^8 \text{ m/s}\).
\(B = \frac{6.3 \text{ V/m}}{3 \times 10^8 \text{ m/s}}\)
\(B = \frac{6.3}{3} \times 10^{-8} \text{ T}\)
\(B = 2.1 \times 10^{-8} \text{ T}\)
Step 2: Determine the Direction of the Magnetic Field
The wave propagates in the positive x-direction (\(\hat{i}\)), so \(\hat{v} = \hat{i}\). The electric field is in the positive y-direction (\(\hat{j}\)), so \(\hat{E} = \hat{j}\).
Using the directional relationship \(\hat{v} = \hat{E} \times \hat{B}\), we have:
\(\hat{i} = \hat{j} \times \hat{B}\)
We need to find the unit vector \(\hat{B}\) such that the cross product of \(\hat{j}\) and \(\hat{B}\) results in \(\hat{i}\). Recall the cyclic properties of the unit vectors in a right-handed coordinate system:
- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{k} = \hat{i}\)
- \(\hat{k} \times \hat{i} = \hat{j}\)
From these relations, we see that \(\hat{j} \times \hat{k} = \hat{i}\). Comparing this with \(\hat{i} = \hat{j} \times \hat{B}\), we conclude that \(\hat{B} = \hat{k}\). The magnetic field is in the positive z-direction.
Step 3: Combine Magnitude and Direction
The magnitude of the magnetic field is \(B = 2.1 \times 10^{-8} \text{ T}\), and its direction is \(\hat{k}\).
Therefore, the magnetic field vector at the given point and time is \(\vec{B} = (2.1 \times 10^{-8} \text{ T}) \hat{k}\).
Comparing with Options
Let's compare our result with the given options:
- \(2.1 \times 10^{-8} \hat{k}\) T
- \(2.1 \times 10^{-6} \hat{j}\) T
- \(1.6 \times 10^{-8} \hat{i}\) T
- \(2.6 \times 10^{-8} \hat{j}\) T
Our calculated magnetic field vector \(\vec{B} = 2.1 \times 10^{-8} \hat{k}\) T matches Option 1.
| Property | Description | Relation |
|---|---|---|
| Speed | Speed of light in free space | \(c \approx 3 \times 10^8\) m/s |
| Field Relation (Magnitude) | Magnitude of E field vs B field | \(E = cB\) |
| Field Relation (Direction) | Direction of propagation, E, and B fields | \(\hat{v} = \hat{E} \times \hat{B}\) |
| Perpendicularity | \(\vec{E}\), \(\vec{B}\), and \(\vec{v}\) are mutually perpendicular | \(\vec{E} \perp \vec{B}\), \(\vec{E} \perp \vec{v}\), \(\vec{B} \perp \vec{v}\) |
Revision Table: Electromagnetic Wave Relations
| Concept | Formula |
|---|---|
| Magnitude Relation | \(E = cB\) or \(B = \frac{E}{c}\) |
| Direction Relation | \(\hat{v} = \hat{E} \times \hat{B}\) |
| Speed of Light in Free Space | \(c \approx 3 \times 10^8\) m/s |
Additional Information on Electromagnetic Waves
Electromagnetic waves are transverse waves produced by the oscillation or acceleration of electric charges. They carry energy through space. In free space, they travel at the speed of light \(c\).
Maxwell's equations are the fundamental equations that describe the behavior of electric and magnetic fields and how they interact, including the generation and propagation of electromagnetic waves. The relationships \(E=cB\) and \(\hat{v} = \hat{E} \times \hat{B}\) are derived from Maxwell's equations.
The frequency (30 MHz in this problem) determines the type of electromagnetic radiation (this is in the radio wave part of the spectrum) and the wavelength (\(\lambda = c/f\)), but it does not affect the fundamental relationship between the magnitudes of the electric and magnetic fields in free space at any given point and time.
The energy flux (power per unit area) carried by an electromagnetic wave is described by the Poynting vector, \(\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})\), where \(\mu_0\) is the permeability of free space. The direction of the Poynting vector is the direction of energy flow, which is also the direction of wave propagation (\(\hat{v}\)). This further confirms that the direction of propagation is given by the cross product \(\hat{E} \times \hat{B}\).
Important Questions from Electromagnetic Waves
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