Question

The electric field in a plane electromagnetic wave is given by \(E_z = 60 \cos(5x + 1.5 \times 10^{10} t)\) V/m. Then expression for the corresponding magnetic field (\(B_y\)) is (here subscripts denote the direction of the field) is:

• A. \( B_y = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^{10} t) \text{ T} \)
• B. \( B_y = 60 \cos(5x + 1.5 \times 10^{10} t) \text{ T} \)
• C. \( B_y = 60 \times 10^9 \cos(5x + 1.5 \times 10^{10} t) \text{ T} \)
• D. \( B_y = -2 \times 10^{-7} \cos(5x + 1.5 \times 10^{10} t) \text{ T} \)

Hint:

Remember the relationship between the amplitudes of electric and magnetic fields in an EM wave \(E_0 = c B_0\), and the direction of propagation is given by \( \vec{E} \times \vec{B} \). Pay attention to the signs and directions.

Answer 1

The Correct Option is D

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

Given the electric field \( E_z = 60 \cos(5x + 1.5 \times 10^{10} t) \, \text{V/m} \), the corresponding magnetic field \(B_y\) is calculated as:

\[ B_y = \frac{E_z}{c} = \frac{60}{3 \times 10^8} \cos(5x + 1.5 \times 10^{10} t) \, \text{T} \]

This simplifies to:

\[ B_y = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^{10} t) \, \text{T} \]

In electromagnetic waves, \(E\), \(B\), and wave propagation are mutually perpendicular and follow the right-hand rule. With \(E\) in the z-direction, \(B\) in the y-direction, and wave propagation in the x-direction, the right-hand rule dictates a negative sign for \(B_y\):

\[ B_y = -2 \times 10^{-7} \cos(5x + 1.5 \times 10^{10} t) \, \text{T} \]