A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At

Q: Why do we use $E = cB$ and not $B = cE$?

The electric field is significantly stronger than the magnetic field in an EM wave by a factor of $c$. The correct derivation from Maxwell's equations yields $E = cB$.

Q: How do I quickly remember the cross product rule for direction?

Just remember the alphabetical order: $\mathbf{E}$ comes before $\mathbf{B}$, so it's $\mathbf{E} \times \mathbf{B}$ to get the velocity $\mathbf{v}$.

Q: What if the wave was traveling in a medium instead of a vacuum?

In a medium, the speed of light $c$ is replaced by the phase velocity $v = c/n$, where $n$ is the refractive index. The relation becomes $E = vB$.

Q: Why did the powers of 10 cancel out completely?

The speed of light has a factor of $10^8$, and the given magnetic field has a factor of $10^{-8}$. When multiplied, $10^8 \times 10^{-8} = 10^0 = 1$.

Q: Can I use Fleming's Left Hand Rule here?

Fleming's Left Hand Rule is for the magnetic force on a moving charge. For EM waves, it's better to use the right-hand rule for the cross product $\mathbf{E} \times \mathbf{B} = \mathbf{v}$.

Question

A plane electromagnetic wave of frequency $25	ext{ GHz}$ is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $\mathbf{B} = 5 	imes 10^{-8} \hat{j}	ext{ T}$. The corresponding electric field $\mathbf{E}$ is (speed of light, $c = 3 	imes 10^8	ext{ ms}^{-1}$)

Options:

(A) $-1.66 	imes 10^{-16} \hat{i}	ext{ V/m}$

(B) $1.66 	imes 10^{-16} \hat{i}	ext{ V/m}$

(C) $-15 \hat{i}	ext{ V/m}$

(D) $15 \hat{i}	ext{ V/m}$

Sigma Percentile · Accepted Answer

### Visualizing the Electromagnetic Wave

Imagine you are standing in a vast, three-dimensional space. The problem tells us that an electromagnetic wave is propagating through a vacuum along the z-direction. This means our velocity vector, $\mathbf{v}$, is pointing straight up along the z-axis, which we can write as $c\hat{k}$.

At the same time, we are given the magnetic field vector, $\mathbf{B}$. It is oscillating along the y-axis, pointing in the $\hat{j}$ direction.

Take a moment to visualize this. You have a wave traveling forward, and its magnetic component is waving side-to-side. Our goal is to find the missing piece of the puzzle: the electric field, $\mathbf{E}$.

### The Master Equation for Magnitude

Let's break this down into two manageable parts: finding the magnitude and finding the direction.

First, the magnitude. In any electromagnetic wave traveling through a vacuum, the electric and magnetic fields are intimately connected. They dance together in perfect harmony, and their amplitudes are related by a beautifully simple equation:

$E = cB$

This equation tells us that the electric field is $c$ times stronger than the magnetic field, where $c$ is the speed of light.

### Executing the Calculation

Now, let's substitute the values we know. The speed of light, $c$, is a universal constant: $3 	imes 10^8 	ext{ m/s}$. The magnitude of our magnetic field, $B$, is given as $5 	imes 10^{-8} 	ext{ T}$.

Plugging these in, we get:

$E = (3 	imes 10^8) 	imes (5 	imes 10^{-8})$

Look closely at those powers of ten. This is where the math becomes elegant. The $10^8$ and $10^{-8}$ perfectly cancel each other out, leaving us with a simple multiplication:

$E = 3 	imes 5 = 15 	ext{ V/m}$

So, the magnitude of our electric field is $15 	ext{ V/m}$. That was the easy part!

### Decoding the Direction

Now comes the crucial step: determining the direction of the electric field. This is where many students make a silly mistake, so let's be careful.

In an electromagnetic wave, the electric field, the magnetic field, and the direction of propagation are mutually perpendicular. They form a 3D orthogonal system. The rule that binds their directions is given by the cross product:

$\hat{E} 	imes \hat{B} = \hat{v}$

We know the direction of the magnetic field is $\hat{j}$, and the direction of propagation is $\hat{k}$. Substituting these into our rule, we get:

$\hat{E} 	imes \hat{j} = \hat{k}$

Now, ask yourself: which unit vector, when crossed with $\hat{j}$, gives $\hat{k}$?

Recall the standard right-hand rule for unit vectors. We know that $\hat{i} 	imes \hat{j} = \hat{k}$.

Therefore, our electric field must be pointing in the $\hat{i}$ direction!

### The Final Conclusion

We have successfully found both pieces of the puzzle. The magnitude is $15$, and the direction is $\hat{i}$.

Combining them, we write our final electric field vector:

$\mathbf{E} = 15\hat{i} 	ext{ V/m}$

This perfectly matches option (d).

By breaking the problem down into magnitude and direction, and relying on the fundamental properties of electromagnetic waves, we navigated through the physics with clarity and confidence. Always remember this orthogonal relationship—it is a favorite concept in JEE!

Select Answer:

Visualized Solution

$Visualizing the Vectors$

$Magnitude Relationship$

$Substituting Values$

$Calculating Magnitude$

$Direction Rule$

$Finding \hat{E}$

$Final Vector$

$Medium Variations$

The Sigma Insight: Characteristics of Electromagnetic Waves

Visualizing the Electromagnetic Wave

The Master Equation for Magnitude

Executing the Calculation

Decoding the Direction

The Final Conclusion

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Visualized Solution

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

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

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

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

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

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

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

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