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

In a plane electromagnetic wave, the electric field \(E\) and magnetic field \(B\) are related through the equation:

\[ B = \frac{E}{c} \]

where \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \text{ m/s}\).

Given the electric field:

\[ E_z = 60 \cos(5x + 1.5 \times 10^{10} t) \, \text{V/m} \]

To find the corresponding magnetic field \(B_y\), we use the relationship:

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

Simplifying:

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

In EM waves, the directions of \(E\), \(B\), and the direction of wave propagation are mutually perpendicular and follow the right-hand rule. For the given configuration, the electric field is in the z-direction, the magnetic field is in the y-direction, and the wave is propagating in the x-direction. Thus, the correct sign for the magnetic field should be negative to adhere to the right-hand rule:

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