The electric field of a plane polarised electromagnetic wave in free space at time t = 0 is given by

Q: Why is the time-dependent phase negative, like $- 10ct$?

Because the wave is propagating in the positive direction of the wave vector $\mathbf{k}$. The standard form for a forward-traveling wave is $\cos(\mathbf{k} \cdot \mathbf{r} - \omega t)$.

Q: How did we know the direction of propagation is $\hat{\mathbf{k}}$?

The spatial part of the phase is $\mathbf{k} \cdot \mathbf{r} = 6x + 8z$. This directly tells us that the wave vector $\mathbf{k}$ is $6\hat{\mathbf{i}} + 8\hat{\mathbf{k}}$. The direction of propagation is simply the unit vector of $\mathbf{k}$.

Q: Why do we divide by 10 when finding $\hat{\mathbf{B}}$?

To find the direction of the magnetic field, we need the *unit* vector of propagation, $\hat{\mathbf{k}}$. Since the magnitude of our wave vector $\mathbf{k} = 6\hat{\mathbf{i}} + 8\hat{\mathbf{k}}$ is 10, we must divide by 10 to normalize it.

Q: Can I use the right-hand rule instead of the cross product?

Absolutely! If you point your right thumb in the direction of propagation ($\mathbf{k}$) and your fingers in the direction of the electric field ($\mathbf{E}$), your palm will push in the direction of the magnetic field ($\mathbf{B}$). However, the cross product is mathematically foolproof.

Q: Why is the amplitude of the magnetic field $10/c$?

In free space, the amplitudes of the electric and magnetic fields are related by the speed of light: $E_0 = c B_0$. Since our given electric field amplitude $E_0$ is 10, $B_0$ must be $10/c$.

Question

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

Options:

(A) $\frac{1}{c}(6\hat{\mathbf{k}} - 8\hat{\mathbf{i}}) \cos[(6x + 8z + 10ct)]$

(B) $\frac{1}{c}(6\hat{\mathbf{k}} - 8\hat{\mathbf{i}}) \cos[(6x + 8z - 10ct)]$

(C) $\frac{1}{c}(6\hat{\mathbf{k}} + 8\hat{\mathbf{i}}) \cos[(6x - 8z + 10ct)]$

(D) $\frac{1}{c}(6\hat{\mathbf{k}} + 8\hat{\mathbf{i}}) \cos[(6x + 8z - 10ct)]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in free space, observing an electromagnetic wave passing by. The problem gives us the electric field of this wave: $\mathbf{E}(x, y) = 10\hat{\mathbf{j}} \cos(6x + 8z)$.

This single equation is packed with information. First, the amplitude of the electric field, $E_0$, is $10$.

Second, the $\hat{\mathbf{j}}$ tells us that the electric field is oscillating strictly along the y-axis.

But the real treasure lies inside the cosine function. The term $6x + 8z$ is the spatial phase of the wave, which dictates how the wave is oriented in space.

### Extracting the Wave Vector

From the spatial phase $6x + 8z$, we can immediately write down the wave vector, $\mathbf{k}$.

The coefficients of $x$ and $z$ give us the components of this vector. So, $\mathbf{k} = 6\hat{\mathbf{i}} + 8\hat{\mathbf{k}}$.

This vector points exactly in the direction the wave is traveling. Let's find its magnitude, $k$.

Using the Pythagorean theorem, $k = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.

Now, we know that for any electromagnetic wave in free space, the angular frequency $\omega$ is related to the wave number $k$ by the speed of light $c$.

**The master relation is $\omega = kc$.**

Substituting our value of $k$, we get $\omega = 10c$.

With this, we can construct the complete time-dependent phase of the wave. Since it's a forward-traveling wave, the phase is $\mathbf{k} \cdot \mathbf{r} - \omega t$, which becomes $6x + 8z - 10ct$.

### The Magnetic Field Direction

Now comes the most thrilling part: finding the magnetic field!

In an electromagnetic wave, the electric field $\mathbf{E}$, the magnetic field $\mathbf{B}$, and the direction of propagation $\mathbf{k}$ are locked in a perfectly perpendicular dance.

**The direction of the magnetic field is always given by the cross product: $\hat{\mathbf{B}} = \hat{\mathbf{k}} 	imes \hat{\mathbf{E}}$.**

Let's set up this cross product carefully. The unit vector of propagation is $\hat{\mathbf{k}} = \frac{6\hat{\mathbf{i}} + 8\hat{\mathbf{k}}}{10}$. The direction of the electric field is $\hat{\mathbf{E}} = \hat{\mathbf{j}}$.

So, $\hat{\mathbf{B}} = \left( \frac{6\hat{\mathbf{i}} + 8\hat{\mathbf{k}}}{10} ight) 	imes \hat{\mathbf{j}}$.

Distributing the cross product, we use the standard right-hand rule for unit vectors: $\hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \hat{\mathbf{k}}$ and $\hat{\mathbf{k}} 	imes \hat{\mathbf{j}} = -\hat{\mathbf{i}}$.

This gives us $\hat{\mathbf{B}} = \frac{6\hat{\mathbf{k}} - 8\hat{\mathbf{i}}}{10}$.

### Final Calculation

We have the direction, and we have the phase. All we need now is the amplitude of the magnetic field, $B_0$.

**In free space, the amplitudes are related by $E_0 = c B_0$.**

Since $E_0 = 10$, we find that $B_0 = \frac{10}{c}$.

Finally, let's assemble the complete magnetic field vector. We multiply the amplitude, the direction unit vector, and the cosine phase together.

$\mathbf{B} = B_0 \hat{\mathbf{B}} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)$

$\mathbf{B} = \frac{10}{c} \left( \frac{6\hat{\mathbf{k}} - 8\hat{\mathbf{i}}}{10} ight) \cos(6x + 8z - 10ct)$

The $10$ in the numerator and denominator beautifully cancel out, leaving us with our final, elegant expression:

$\mathbf{B} = \frac{1}{c}(6\hat{\mathbf{k}} - 8\hat{\mathbf{i}}) \cos(6x + 8z - 10ct)$

And there we have it! A perfect match with option (b).

Animated Solution for Physics - Electromagnetic Waves: 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)

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Visualized Solution

$Analyzing the Electric Field$

$Extracting the Wave Vector$

$Calculating Angular Frequency$

$Constructing the Full Phase$

$Direction of Magnetic Field$

$Evaluating the Cross Product$

$Amplitude of Magnetic Field$

$Final Magnetic Field Equation$

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup

Extracting the Wave Vector

The Magnetic Field Direction

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting the Wave Vector

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constructing the Full Phase

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction of Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Cross Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Amplitude of Magnetic Field

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