Question:

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time), \( \vec{E} = 9.6 \hat{j} \, \text{V/m} \). The value of the magnetic field at this point is:

Updated On: Nov 3, 2025
  • \( 3.2 \times 10^{-8} \hat{k} \, \text{T} \)
  • \( 3.2 \times 10^{-8} \hat{i} \, \text{T} \)
  • \( 9.6 \hat{j} \, \text{T} \)
  • \( 9.6 \times 10^{-8} \hat{k} \, \text{T} \)
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The Correct Option is A

Approach Solution - 1

To find the magnetic field corresponding to the given electric field of an electromagnetic wave, we use the relationship between the electric and magnetic fields in a plane electromagnetic wave. The relationship is given by:

\(E = cB\)

where:

  • \(E\) is the magnitude of the electric field,
  • \(B\) is the magnitude of the magnetic field,
  • \(c\) is the speed of light in free space, approximately \(3 \times 10^8 \text{ m/s}\).

Given:

  • Electric field, \(\vec{E} = 9.6 \hat{j} \, \text{V/m}\)

Since the wave is traveling in the X-direction, the magnetic field will be perpendicular to both the direction of wave propagation (X-direction) and the electric field (Y-direction). Therefore, it must be in the Z-direction. Let's calculate the magnitude of the magnetic field:

\(B = \frac{E}{c} = \frac{9.6}{3 \times 10^8}\)

Calculating the value:

\(B = 3.2 \times 10^{-8} \, \text{T}\)

Since the magnetic field is in the Z-direction, we write the magnetic field as:

\(\vec{B} = 3.2 \times 10^{-8} \hat{k} \, \text{T}\)

Thus, the correct option is:

\(3.2 \times 10^{-8} \hat{k} \, \text{T}\)

Let's rule out other options:

  • \(3.2 \times 10^{-8} \hat{i} \, \text{T}\): Incorrect, since it implies the magnetic field is in the X-direction, which cannot be the case as the wave travels in the X-direction.
  • \(9.6 \hat{j} \, \text{T}\): Incorrect, as it suggests a mismatched unit and direction against the derived formula.
  • \(9.6 \times 10^{-8} \hat{k} \, \text{T}\): Incorrect, wrong magnitude of the magnetic field.

Therefore, the value of the magnetic field at this point is \(3.2 \times 10^{-8} \hat{k} \, \text{T}\).

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Approach Solution -2

For an electromagnetic wave, the electric field \( \vec{E} \) and magnetic field \( \vec{B} \) are related by the speed of light \( c \):  
\(\frac{E}{B} = c\)
where \( c = 3 \times 10^8 \, \text{m/s} \).

Step 1: Calculate the magnetic field \( B \):  
Given \( E = 9.6 \, \text{V/m} \),  
\(B = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8} \, \text{T}\)

Step 2: Determine the direction of \( \vec{B} \):  
Since the wave travels along the X-direction and \( \vec{E} \) is along \( \hat{j} \), the magnetic field \( \vec{B} \) must be perpendicular to both the direction of propagation and \( \vec{E} \). By the right-hand rule:  
\(\vec{B} \, \text{points along} \, \hat{k}.\)

Thus, the magnetic field at this point is \( 3.2 \times 10^{-8} \hat{k} \, \text{T}. \)

The Correct Answer is: \( 3.2 \times 10^{-8} \hat{k} \, \text{T}. \)

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