The electric field of a plane electromagnetic wave is given by E = E_0 i cos(kz) cos(wt). The corres

Q: Why did we use Faraday's law instead of just $\mathbf{B} = \frac{1}{c} (\hat{\mathbf{k}} \times \mathbf{E})$?

The formula $\mathbf{B} = \frac{1}{c} (\hat{\mathbf{k}} \times \mathbf{E})$ is strictly valid for a single traveling plane wave (like $\cos(kz - \omega t)$). The given electric field $\cos(kz)\cos(\omega t)$ represents a standing wave, which is a superposition of two traveling waves. Using Faraday's law ($\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$) is a fundamental approach that works for any electromagnetic field configuration.

Q: What happens to the constant of integration when we integrate $\frac{\partial \mathbf{B}}{\partial t}$?

When we integrate to find the magnetic field, we could technically add a constant of integration (a static magnetic field). However, in the context of electromagnetic waves, we are interested in the oscillating components. A static field doesn't contribute to the wave propagation, so we set the constant to zero.

Q: Why is the derivative of $\cos(kz)$ with respect to $z$ equal to $-k\sin(kz)$?

This is an application of the chain rule in calculus. The derivative of the outer function $\cos(u)$ is $-\sin(u)$, and we multiply it by the derivative of the inner function $u = kz$ with respect to $z$, which is $k$.

Q: How do we know the direction of the magnetic field?

The curl operation naturally gives us the direction. When we computed $\nabla \times \mathbf{E}$, the non-zero term was associated with the $\hat{\mathbf{j}}$ unit vector. This mathematically confirms that the oscillating magnetic field is directed along the y-axis.

Q: What does the term $c = \omega / k$ represent?

Here, $\omega$ is the angular frequency (how fast the wave oscillates in time) and $k$ is the wave number (how fast it oscillates in space). Their ratio $\omega / k$ gives the phase velocity of the wave, which for an electromagnetic wave in a vacuum is the speed of light, $c$.

Question

The electric field of a plane electromagnetic wave is given by
$\mathbf{E} = E_0\hat{\mathbf{i}}\cos(kz)\cos(\omega t)$
The corresponding magnetic field $\mathbf{B}$ is then given by

Options:

(A) $\mathbf{B} = \frac{E_0}{C}\hat{\mathbf{j}}\sin(kz)\sin(\omega t)$

(B) $\mathbf{B} = \frac{E_0}{C}\hat{\mathbf{j}}\sin(kz)\cos(\omega t)$

(C) $\mathbf{B} = \frac{E_0}{C}\hat{\mathbf{k}}\sin(kz)\cos(\omega t)$

(D) $\mathbf{B} = \frac{E_0}{C}\hat{\mathbf{j}}\cos(kz)\sin(\omega t)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing an electromagnetic wave, but it's not your standard traveling wave. The electric field is given by:
$$ \mathbf{E} = E_0 \hat{\mathbf{i}} \cos(kz) \cos(\omega t) $$

Notice the product of two cosine functions—one depending on space ($z$) and the other on time ($t$). This specific mathematical form represents a **standing wave**, which is essentially the superposition of two waves traveling in opposite directions. Because it's not a simple traveling wave of the form $\cos(kz - \omega t)$, we cannot just use the shortcut formula $\mathbf{B} = \frac{1}{c} (\hat{\mathbf{k}} 	imes \mathbf{E})$.

Instead, we must rely on the fundamental laws of electromagnetism. We need a tool that connects the spatial variation of the electric field to the time variation of the magnetic field.

### The Master Equation: Faraday's Law

To find the magnetic field $\mathbf{B}$, we turn to Maxwell's equations, specifically **Faraday's Law of Induction** in its differential form:
$$ 
abla 	imes \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$

This elegant equation tells us that the curl (spatial rotation) of the electric field generates a time-varying magnetic field. Let's compute the curl of our given electric field. We set up the determinant:
$$ 
abla 	imes \mathbf{E} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ E_x & 0 & 0 \end{vmatrix} $$

Since our electric field only has an $x$-component ($E_x$) and it only depends on $z$ and $t$, the partial derivatives with respect to $x$ and $y$ are zero. Expanding the determinant, we are left with just one non-zero term:
$$ 
abla 	imes \mathbf{E} = \hat{\mathbf{j}} \frac{\partial E_x}{\partial z} $$

### Executing the Calculus

Now, let's substitute our $E_x$ into the derivative:
$$ \frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z} [E_0 \cos(kz) \cos(\omega t)] $$

Applying the chain rule, the derivative of $\cos(kz)$ is $-k \sin(kz)$. The time-dependent part $\cos(\omega t)$ acts as a constant here.
$$ 
abla 	imes \mathbf{E} = -k E_0 \sin(kz) \cos(\omega t) \hat{\mathbf{j}} $$

Equating this to the right side of Faraday's Law:
$$ -\frac{\partial \mathbf{B}}{\partial t} = -k E_0 \sin(kz) \cos(\omega t) \hat{\mathbf{j}} $$

The negative signs cancel out beautifully:
$$ \frac{\partial \mathbf{B}}{\partial t} = k E_0 \sin(kz) \cos(\omega t) \hat{\mathbf{j}} $$

### Final Calculation and Integration

We have the rate of change of the magnetic field. To find the actual magnetic field $\mathbf{B}$, we integrate with respect to time $t$:
$$ \mathbf{B} = \int k E_0 \sin(kz) \cos(\omega t) \hat{\mathbf{j}} \, dt $$

The integral of $\cos(\omega t)$ is $\frac{\sin(\omega t)}{\omega}$.
$$ \mathbf{B} = k E_0 \sin(kz) \left( \frac{\sin(\omega t)}{\omega} ight) \hat{\mathbf{j}} $$

Finally, we use the fundamental wave relationship connecting the wave number $k$, angular frequency $\omega$, and the speed of light $c$:
$$ c = \frac{\omega}{k} \implies \frac{k}{\omega} = \frac{1}{c} $$

Substituting this into our expression, we arrive at the final magnetic field:
$$ \mathbf{B} = \frac{E_0}{c} \sin(kz) \sin(\omega t) \hat{\mathbf{j}} $$

This perfectly matches option (a), confirming our rigorous mathematical journey from Maxwell's equations to the final wave function!

Animated Solution for Physics - Electromagnetic Waves: 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

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Visualized Solution

$The Given Electric Field$

$Faraday’s Law of Induction$

$Calculating the Curl of E$

$Differentiating E$

$Equating to - \frac{\partial B}{\partial t}$

$Integrating to find B$

$Final Simplification$

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup

The Master Equation

Executing the Calculus

Final Calculation and Integration

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Given Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday’s Law of Induction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Curl of E

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating E

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating to −∂t∂B​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating to find B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Simplification