In a plane electromagnetic wave, the directions of electric field and magnetic field are represented

Q: Why do we use $\vec{E} \times \vec{B}$ and not $\vec{B} \times \vec{E}$?

The direction of propagation is derived from the Poynting vector $\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$. Using $\vec{B} \times \vec{E}$ would give the exact opposite direction, which violates the right-hand rule for electromagnetic waves.

Q: Can I just guess the direction since it has to be perpendicular to $\hat{k}$?

While you know it must lie in the $xy$-plane (since it's perpendicular to $\hat{k}$), there are infinitely many directions in that plane. The cross product gives you the exact unique direction.

Q: I always mess up $\hat{k} \times \hat{j}$. How do I remember it?

Use the cyclic circle: $\hat{i} \rightarrow \hat{j} \rightarrow \hat{k} \rightarrow \hat{i}$. Going clockwise (alphabetical order) gives positive results ($\hat{i} \times \hat{j} = \hat{k}$). Going counter-clockwise gives negative results ($\hat{k} \times \hat{j} = -\hat{i}$).

Q: Why did we ignore the magnitudes of the actual fields and just use their direction vectors?

The question only asks for the *direction* of propagation. The magnitudes of $\vec{E}$ and $\vec{B}$ affect the amplitude of the wave and the energy it carries, but the geometric direction is purely determined by their unit vectors.

Q: What if the magnetic field had a $\hat{k}$ component?

In a plane electromagnetic wave in a vacuum, $\vec{E}$ and $\vec{B}$ are always strictly perpendicular. If $\vec{E}$ is along $\hat{k}$, $\vec{B}$ cannot have a $\hat{k}$ component. If it did, their dot product wouldn't be zero!

Question

In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by $\hat{k}$ and $2\hat{i} - 2\hat{j}$, respectively. What is the unit vector along direction of propagation of the wave?

Options:

(A) $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$

(B) $\frac{1}{\sqrt{2}}(\hat{j} + \hat{k})$

(C) $\frac{1}{\sqrt{5}}(\hat{i} + 2\hat{j})$

(D) $\frac{1}{\sqrt{5}}(2\hat{i} + \hat{j})$

Sigma Percentile · Accepted Answer

The journey of an electromagnetic wave through space is a beautiful dance of geometry and physics. Imagine you are standing in a three-dimensional coordinate system. The electric field ($\vec{E}$) and the magnetic field ($\vec{B}$) are oscillating, but they aren't just flailing around randomly. They follow a strict, elegant rule: they must always be perpendicular to each other, and the wave itself must travel in a direction perpendicular to both of them.

In this problem, we are given the exact directions of these fields. The electric field is pointing straight up along the z-axis, represented by the unit vector $\hat{k}$. Meanwhile, the magnetic field is sweeping across the xy-plane, pointing in the direction of $2\hat{i} - 2\hat{j}$.

Our mission? To find the exact direction the wave is traveling.

### The Master Equation

To find the direction of propagation, we rely on a fundamental property of electromagnetic waves. The direction in which the wave carries energy is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic fields.

Therefore, the direction of propagation, let's call it $\vec{v}$, is parallel to $\vec{E} 	imes \vec{B}$.

**This is a favorite concept for JEE.** The order of the cross product is absolutely critical. It must be $\vec{E} 	imes \vec{B}$. If you accidentally calculate $\vec{B} 	imes \vec{E}$, you will find the wave traveling backwards!

### Setting Up the Cross Product

Let's substitute the given vectors into our cross product setup. We know that $\vec{E}$ is along $\hat{k}$ and $\vec{B}$ is along $2\hat{i} - 2\hat{j}$.

$$ \vec{v} \parallel \hat{k} 	imes (2\hat{i} - 2\hat{j}) $$

Now, we can use the distributive property of the cross product to expand this expression. We can pull the constant $2$ out to make things cleaner.

$$ \vec{v} \parallel 2(\hat{k} 	imes \hat{i}) - 2(\hat{k} 	imes \hat{j}) $$

### Navigating the Unit Vectors

This is where mistakes happen. We need to evaluate the cross products of the fundamental unit vectors. To do this safely, always remember the cyclic circle: $\hat{i} ightarrow \hat{j} ightarrow \hat{k} ightarrow \hat{i}$.

When we multiply $\hat{k} 	imes \hat{i}$, we are moving forward in the cycle, so the result is positive $\hat{j}$.

However, when we multiply $\hat{k} 	imes \hat{j}$, we are moving backward against the cycle. Therefore, the result must be negative, giving us $-\hat{i}$.

Let's substitute these results back into our equation. Watch out for the minus sign!

$$ \vec{v} \parallel 2(\hat{j}) - 2(-\hat{i}) $$

The two negative signs cancel each other out, leaving us with a beautiful, symmetric vector.

$$ \vec{v} \parallel 2\hat{i} + 2\hat{j} $$

### The Final Polish

We have found the direction of propagation! The wave is traveling along the vector $2\hat{i} + 2\hat{j}$, which means it is moving diagonally across the xy-plane.

But we aren't quite done. The question specifically asks for the **unit vector** along this direction. To find the unit vector, we must divide our vector by its own magnitude.

First, let's calculate the magnitude of $\vec{v}$.

$$ |\vec{v}| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

Now, we divide the vector by this magnitude to normalize it.

$$ \hat{v} = \frac{2\hat{i} + 2\hat{j}}{2\sqrt{2}} $$

We can factor out a $2$ from the numerator and cancel it with the $2$ in the denominator.

$$ \hat{v} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j}) $$

And there we have it! The unit vector pointing in the direction of the wave's propagation is $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$. This elegant result shows that the wave is slicing perfectly at a 45-degree angle between the x and y axes.

Animated Solution for Physics - Electromagnetic Waves: In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by $\hat{k}$ and $2 \hat{i} - 2 \hat{j}$ , respectively. What is the unit vector along direction of propagation of the wave?

Select Answer:

Visualized Solution

Visualizing the Vectors

Direction of Propagation

Setting up the Cross Product

Distributing the Cross Product

Unit Vector Cross Products

Resulting Vector

Finding the Magnitude

The Unit Vector

Conclusion

The Sigma Insight: Characteristics of Electromagnetic Waves

The Master Equation

Setting Up the Cross Product

Navigating the Unit Vectors

The Final Polish

Similar Questions

An electromagnetic wave is represented by the electric field $E = E_{0} \hat{n} sin [ω t + (6 y - 8 z)]$ . Taking unit vectors in $x$ , $y$ and $z$ -directions to be $\hat{i}, \hat{j}, \hat{k}$ , the direction of propagation $\hat{s}$ , is

A plane electromagnetic wave is propagating along the direction $\frac{i ^ + j ^}{2}$ with its polarisation along the direction $\hat{k}$ . The correct form of the magnetic field of the wave would be (here $B_{0}$ is an appropriate constant)

In an electromagnetic wave, the electric field vector and magnetic field vector are given as $E = E_{0} \hat{i}$ and $B = B_{0} \hat{k}$ , respectively. The direction of propagation of electromagnetic wave is along

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

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

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

A plane electromagnetic wave of frequency $100 MHz$ is travelling in vacuum along the x-direction. At a particular point in space and time, $B = 2.0 \times 1 0^{- 8} \hat{k} T$ (where, $\hat{k}$ is unit vector along z-direction). What is $E$ at this point?

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

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)