The magnetic field of a plane electromagnetic wave is B = 3 x 10^-8 sin[200pi(y + ct)] i T, where c

Q: Why do we differentiate the phase to find the velocity?

The phase of a wave represents a specific point on the waveform (like a crest or trough). By setting the phase to a constant and differentiating with respect to time, we are tracking how fast that specific point moves through space, which gives us the wave velocity.

Q: How did we know the magnetic field is in the $\hat{i}$ direction?

The problem explicitly states the magnetic field vector as $\mathbf{B} = B_0 \sin(\dots) \hat{i}$. The $\hat{i}$ at the end of the expression is the unit vector indicating the direction of the magnetic field oscillation.

Q: What if the phase was $ky - \omega t$?

If the phase was $ky - \omega t$, setting it to a constant and differentiating would give $k \frac{dy}{dt} - \omega = 0$, or $\frac{dy}{dt} = \frac{\omega}{k} = c$. This would mean the wave is traveling in the positive y-direction ($+\hat{j}$).

Q: Why is the direction of propagation $\hat{E} \times \hat{B}$ and not $\hat{B} \times \hat{E}$?

This comes directly from Maxwell's equations and the Poynting vector, which describes the directional energy flux of an electromagnetic field. The Poynting vector is defined as $\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$, so the energy (and the wave itself) flows in the direction of $\mathbf{E} \times \mathbf{B}$.

Q: Can the electric and magnetic fields have different phases?

In a plane electromagnetic wave traveling in a vacuum or a lossless dielectric medium, the electric and magnetic fields are always perfectly in phase. They reach their maximum and minimum values at the exact same time and position.

Question

The magnetic field of a plane electromagnetic wave is
$\mathbf{B} = 3 	imes 10^{-8} \sin[200\pi(y + ct)] \hat{i}$ T,
where $c = 3 	imes 10^8$ ms$^{-1}$ is the speed of light. The corresponding electric field is

Options:

(A) $\mathbf{E} = 9 \sin[200\pi(y + ct)] \hat{k}$ V/m

(B) $\mathbf{E} = -10^{-6} \sin[200\pi(y + ct)] \hat{k}$ V/m

(C) $\mathbf{E} = 3 	imes 10^{-8} \sin[200\pi(y + ct)] \hat{k}$ V/m

(D) $\mathbf{E} = -9 \sin[200\pi(y + ct)] \hat{k}$ V/m

Sigma Percentile · Accepted Answer

The journey of mastering Electromagnetic Waves often brings us face-to-face with the elegant interplay between electric and magnetic fields. In this problem, we are handed the magnetic field of a plane electromagnetic wave and tasked with uncovering its electric field counterpart.

I know this might look like a dense mathematical expression at first glance, but let's take a breath. By breaking it down into its core physical principles, we will see just how beautifully symmetric nature truly is.

### Decoding the Magnetic Field

Let's start by looking closely at the given magnetic field equation:
$$ \mathbf{B} = 3 	imes 10^{-8} \sin[200\pi(y + ct)] \hat{i} 	ext{ T} $$

This single line of math is packed with physical reality. The coefficient $3 	imes 10^{-8}$ is our amplitude, $B_0$. The sine function describes the wave's oscillation through space and time. Finally, the $\hat{i}$ vector at the very end tells us that the magnetic field is oscillating strictly along the positive x-axis.

### The Amplitude Connection

Our first major goal is to find the amplitude of the electric field, $E_0$. In a vacuum, the electric and magnetic fields of an electromagnetic wave are intimately connected by the speed of light, $c$. The relationship is beautifully simple:
$$ E_0 = c B_0 $$

Let's substitute the values we know. The speed of light is $3 	imes 10^8 	ext{ m/s}$, and our magnetic field amplitude is $3 	imes 10^{-8} 	ext{ T}$.
$$ E_0 = (3 	imes 10^8) 	imes (3 	imes 10^{-8}) $$

Notice how perfectly the powers of ten cancel each other out! This leaves us with a clean, whole number:
$$ E_0 = 9 	ext{ V/m} $$

### Unmasking the Direction of Propagation

Now that we have the amplitude, we need to figure out which way this wave is actually traveling. To do this, we look inside the sine function at the **phase** of the wave: $200\pi(y + ct)$.

Imagine you are surfing on a specific crest of this wave. For you, the phase remains constant. If we set the phase to a constant and differentiate it with respect to time, we can find your velocity!
$$ y + ct = 	ext{constant} $$
$$ \frac{dy}{dt} + c = 0 \implies \frac{dy}{dt} = -c $$

The negative sign is the crucial detail here. It tells us that the wave is propagating in the negative y-direction. In vector notation, the direction of propagation is $\hat{v} = -\hat{j}$.

### The Cross Product Puzzle

We have the amplitude, and we know the wave is moving along $-\hat{j}$. We also know the magnetic field oscillates along $\hat{i}$. How do we find the direction of the electric field?

This is where the fundamental geometry of electromagnetic waves comes into play. The direction of propagation is always given by the cross product of the electric and magnetic field unit vectors:
$$ \hat{v} = \hat{E} 	imes \hat{B} $$

Let's plug in what we know:
$$ -\hat{j} = \hat{E} 	imes \hat{i} $$

Now, we have to solve a small vector puzzle. What unit vector, when crossed with $\hat{i}$, gives $-\hat{j}$? We know from standard vector math that $\hat{k} 	imes \hat{i} = \hat{j}$. Therefore, to get a negative $\hat{j}$, we must use $-\hat{k}$.
$$ (-\hat{k}) 	imes \hat{i} = -\hat{j} $$

So, the electric field must oscillate along the negative z-axis, or $-\hat{k}$.

### The Final Masterpiece

We have successfully gathered all the pieces of the puzzle. We have the amplitude ($9 	ext{ V/m}$), the phase ($200\pi(y + ct)$), and the direction ($-\hat{k}$).

Let's assemble the final electric field vector:
$$ \mathbf{E} = E_0 \sin(	ext{phase}) \hat{E} $$
$$ \mathbf{E} = 9 \sin[200\pi(y + ct)] (-\hat{k}) $$
$$ \mathbf{E} = -9 \sin[200\pi(y + ct)] \hat{k} 	ext{ V/m} $$

And there we have it! By systematically applying the core principles of electromagnetic waves, we've translated the magnetic field into its corresponding electric field.

Animated Solution for Physics - Electromagnetic Waves: The magnetic field of a plane electromagnetic wave is $B = 3 \times 1 0^{- 8} sin [200 π (y + c t)] \hat{i}$ T, where $c = 3 \times 1 0^{8}$ ms $^{- 1}$ is the speed of light. The corresponding electric field is

Select Answer:

Visualized Solution

Visualizing the Magnetic Field

Amplitude Relationship

Substituting Values

Calculating Electric Field Amplitude

Analyzing the Phase

Direction of Propagation

The Cross Product Rule

Setting Up the Cross Product

Solving for Electric Field Direction

Final Electric Field Vector

The Way Forward

The Sigma Insight: Characteristics of Electromagnetic Waves

Decoding the Magnetic Field

The Amplitude Connection

Unmasking the Direction of Propagation

The Cross Product Puzzle

The Final Masterpiece

Similar Questions

The electric field of a plane electromagnetic wave is given by $E = E_{0} (\hat{x} + \hat{y}) sin (k z - ω t)$ Its magnetic field will be given by

The electric field of a plane electromagnetic wave is given by $E = E_{0} \hat{i} cos (k z) cos (ω t)$ The corresponding magnetic field $B$ is then given by

The electric field of a plane electromagnetic wave propagating along the x-direction in vacuum is $E = E_{0} \hat{j} cos (ω t - k x)$ . The magnetic field $B$ , at the moment $t = 0$ is

For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is $B (x, t) = [1.2 \times 1 0^{- 7} sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{k}] T$ . The instantaneous electric field $E$ corresponding to $B$ is (Given, speed of light, $c = 3 \times 1 0^{8} m/s$ )

The electric field of a plane polarised electromagnetic wave in free space at time $t = 0$ is given by an expression. $E (x, y) = 10 \hat{j} cos [(6 x + 8 z)]$ The magnetic field $B (x, z, t)$ is given by (where, $c$ is the velocity of light)

If the magnetic field of a plane electromagnetic wave is given by $B = 100 \times 1 0^{- 6} sin [2 π \times 2 \times 1 0^{15} (t - \frac{x}{c})]$ then the maximum electric field associated with it is (Take, the speed of light $= 3 \times 1 0^{8}$ m/s)

A plane electromagnetic wave of frequency $50 MHz$ travels in free space along the positive $x$ -direction. At a particular point in space and time, $E = 6.3 \hat{j} V/m$ . The corresponding magnetic field $B$ , at that point will be

A plane electromagnetic wave of frequency $25 GHz$ is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $B = 5 \times 1 0^{- 8} \hat{j} T$ . The corresponding electric field $E$ is (speed of light, $c = 3 \times 1 0^{8} ms^{- 1}$ )

The magnetic field of an electromagnetic wave is given by $B = 1.6 \times 1 0^{- 6} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (2 \hat{i} + \hat{j}) Wbm^{- 2}$ The associated electric field will be

A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is $E = 6 Vm^{- 1}$ along y-direction. Its corresponding magnetic field component, $B$ would be