Question

A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be \(x \times 10^{-8}\) T. The value of x is _________.

Hint:

Remember the simple relation \(E = cB\) for EM waves in vacuum. It's a fundamental concept. Note that the electric field value in V/m is always a much larger number than the magnetic field value in Tesla. This is because of the large value of \(c\). Be careful not to use the formula for a medium, \(E=vB\), unless specified.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
We are given the magnitude of the electric field of an electromagnetic wave in free space and need to find the magnitude of the magnetic field.
Step 2: Key Formula or Approach:
In an electromagnetic wave traveling in a vacuum (free space), the ratio of the magnitudes of the electric field (E) and the magnetic field (B) at any instant is equal to the speed of light in vacuum (c).
\[ \frac{E}{B} = c \] where \( c \approx 3 \times 10^8 \) m/s.
Step 3: Detailed Explanation:
Given values:
- Electric field magnitude, \( E = 6 \) V/m.
- Speed of light, \( c = 3 \times 10^8 \) m/s.
The frequency of the wave (30 MHz) is extra information and not needed for this calculation.
Using the formula, we can solve for the magnetic field magnitude B:
\[ B = \frac{E}{c} \] \[ B = \frac{6 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} = 2 \times 10^{-8} \, \text{T} \] The problem states that the magnetic field is \( x \times 10^{-8} \) T.
By comparing our result with the given expression, we find:
\[ x = 2 \] Step 4: Final Answer:
The value of x is 2.