Question

If the electric field of a plane electromagnetic wave is $E_z = 60\sin(0.5\times 10^3x + 1.5\times 10^{11}t)$ Vm$^{-1}$, then the magnetic field of the wave is

• A. $B_y = 2\times 10^{-7}\sin(0.5\times 10^3x + 1.5\times 10^{11}t)$ T
• B. $B_z = 2\times 10^{-7}\sin(0.5\times 10^3x + 1.5\times 10^{11}t)$ T
• C. $B_x = 180\times 10^8\sin(0.5\times 10^3x + 1.5\times 10^{11}t)$ T
• D. $B_y = 180\times 10^8\sin(0.5\times 10^3x + 1.5\times 10^{11}t)$ T

Hint:

In a plane electromagnetic wave, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the direction of propagation ($\vec{v}$). The direction of propagation is given by $\vec{v} \propto \vec{E} \times \vec{B}$, and the ratio of their amplitudes is $E_0/B_0 = c$.

Answer 1

The Correct Option is A

Step 1: Use the relation between the magnitudes of the electric and magnetic fields.
The ratio of the magnitudes of the electric field ($E_0$) and the magnetic field ($B_0$) in a plane electromagnetic wave is equal to the speed of light in a vacuum ($c$). \[ c = \frac{E_0}{B_0} \implies B_0 = \frac{E_0}{c}. \] The speed of light is $c \approx 3 \times 10^8 \text{ m/s}$.

Step 2: Calculate the amplitude of the magnetic field $B_0$.
From the given electric field equation $E_z = 60\sin(kx + \omega t)$, the amplitude is $E_0 = 60 \text{ V/m}$. \[ B_0 = \frac{60 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 20 \times 10^{-8} \text{ T} = 2 \times 10^{-7} \text{ T}. \]

Step 3: Determine the direction of the magnetic field using the rule of propagation.
The direction of propagation of the electromagnetic wave is given by the direction of the vector $\vec{E} \times \vec{B}$. The wave is propagating along the negative x-direction because the term in the argument of the sine function is $kx + \omega t$. The electric field oscillates along the z-axis ($\vec{E} = E_z \hat{k}$). The direction of propagation is $\vec{v} \propto \frac{\partial}{\partial x} (\text{argument}) = k \hat{i}$. Wait, the $\sin(kx+\omega t)$ form means propagation in the negative x-direction. Let's assume the standard $\sin(kx-\omega t)$ means positive x-direction, and $\sin(kx+\omega t)$ means negative x-direction. $\vec{v} \propto -\hat{i}$. We require $\vec{E} \times \vec{B}$ to be in the direction of propagation, i.e., $-\hat{i}$. \[ \vec{E} \times \vec{B} \propto \hat{k} \times \vec{B} = -\hat{i}. \] Since $\hat{k} \times \hat{j} = -\hat{i}$, the magnetic field must be along the y-axis ($\vec{B} = B_y \hat{j}$).

Step 4: Write the final equation for the magnetic field.
The magnetic field must oscillate in phase with the electric field, so the functional form remains the same. \[ B_y = B_0 \sin(0.5\times 10^3x + 1.5\times 10^{11}t) \] \[ B_y = 2\times 10^{-7}\sin(0.5\times 10^3x + 1.5\times 10^{11}t) \text{ T}. \]