Question:
In an EM wave, electric field amplitude is \( 300\,\text{V/m} \). Magnetic field amplitude is:
In an EM wave, electric field amplitude is \( 300\,\text{V/m} \). Magnetic field amplitude is:
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The magnetic field vector in an EM wave is always much smaller in numerical value than the electric field vector because of the factor of $c$ ($3 \times 10^8$) in the denominator.
Always express your final answer in standard scientific notation to easily match the options.
Always express your final answer in standard scientific notation to easily match the options.
Updated On: Jun 3, 2026
- $1 \times 10^{-6}\text{ T}$
- $1 \times 10^{-5}\text{ T}$
- $1 \times 10^{-7}\text{ T}$
- $3 \times 10^{-6}\text{ T}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks to find the amplitude of the magnetic field vector in an electromagnetic (EM) wave, given the amplitude of its corresponding electric field vector.
Step 2: Key Formula or Approach:
In any electromagnetic wave propagating in a vacuum or air, the amplitudes of the electric field ($E_0$) and the magnetic field ($B_0$) are directly related to the speed of light ($c$) by the equation:
\[ c = \frac{E_0}{B_0} \] Rearranging this formula to find the magnetic field amplitude:
\[ B_0 = \frac{E_0}{c} \] where $c \approx 3 \times 10^8\text{ m/s}$ is the speed of light in vacuum.
Step 3: Detailed Explanation:
• Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of wave propagation.
• The ratio of the magnitude of the electric field to the magnitude of the magnetic field at any instant is a constant, which is equal to the velocity of the wave in that medium.
• We are given the electric field amplitude as $E_0 = 300\text{ V/m}$.
• The speed of light in vacuum is a fundamental constant, $c = 3 \times 10^8\text{ m/s}$.
• Substituting these values into the rearranged relation:
\[ B_0 = \frac{300}{3 \times 10^8} \] \[ B_0 = 100 \times 10^{-8}\text{ T} \] \[ B_0 = 1 \times 10^{-6}\text{ T} \]
• Thus, the amplitude of the oscillating magnetic field in this electromagnetic wave is $1 \times 10^{-6}\text{ T}$.
Step 4: Final Answer:
The magnetic field amplitude of the EM wave is $1 \times 10^{-6}\text{ T}$.
The question asks to find the amplitude of the magnetic field vector in an electromagnetic (EM) wave, given the amplitude of its corresponding electric field vector.
Step 2: Key Formula or Approach:
In any electromagnetic wave propagating in a vacuum or air, the amplitudes of the electric field ($E_0$) and the magnetic field ($B_0$) are directly related to the speed of light ($c$) by the equation:
\[ c = \frac{E_0}{B_0} \] Rearranging this formula to find the magnetic field amplitude:
\[ B_0 = \frac{E_0}{c} \] where $c \approx 3 \times 10^8\text{ m/s}$ is the speed of light in vacuum.
Step 3: Detailed Explanation:
• Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of wave propagation.
• The ratio of the magnitude of the electric field to the magnitude of the magnetic field at any instant is a constant, which is equal to the velocity of the wave in that medium.
• We are given the electric field amplitude as $E_0 = 300\text{ V/m}$.
• The speed of light in vacuum is a fundamental constant, $c = 3 \times 10^8\text{ m/s}$.
• Substituting these values into the rearranged relation:
\[ B_0 = \frac{300}{3 \times 10^8} \] \[ B_0 = 100 \times 10^{-8}\text{ T} \] \[ B_0 = 1 \times 10^{-6}\text{ T} \]
• Thus, the amplitude of the oscillating magnetic field in this electromagnetic wave is $1 \times 10^{-6}\text{ T}$.
Step 4: Final Answer:
The magnetic field amplitude of the EM wave is $1 \times 10^{-6}\text{ T}$.
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