A plane electromagnetic wave, has frequency of 2.0 x 10^10 Hz and its energy density is 1.02 x 10^-8

Q: Why didn't we use the frequency given in the question?

In many competitive exams, extra information is provided to test your conceptual clarity. The energy density formula $u = \frac{B_0^2}{2\mu_0}$ depends only on the magnetic field amplitude and the permeability of free space, making the frequency irrelevant for this specific calculation.

Q: Is the given energy density just for the magnetic field or the whole wave?

The problem states "its energy density", referring to the entire electromagnetic wave. The total energy density is the sum of the electric and magnetic energy densities, which are equal in magnitude.

Q: Why is the formula $u = \frac{B_0^2}{2\mu_0}$ and not $\frac{B_0^2}{4\mu_0}$?

The energy density of *just* the magnetic field is $\frac{B_0^2}{4\mu_0}$. Since the total energy density is equally shared between the electric and magnetic fields, the total energy density is $2 \times \frac{B_0^2}{4\mu_0} = \frac{B_0^2}{2\mu_0}$.

Q: How do I quickly calculate the square root of $2.56 \times 10^{-14}$ without a calculator?

Recognize that $256$ is a perfect square ($16^2 = 256$). Therefore, $\sqrt{2.56} = 1.6$. For the power of 10, simply halve the even exponent: $\sqrt{10^{-14}} = 10^{-7}$.

Q: What is the significance of $\mu_0$ in this equation?

$\mu_0$ is the magnetic permeability of free space. It dictates how easily a magnetic field can establish itself in a vacuum, which directly relates the magnetic field's amplitude to the energy it stores.

Question

A plane electromagnetic wave, has frequency of $2.0 	imes 10^{10}$ Hz and its energy density is $1.02 	imes 10^{-8} 	ext{ J/m}^3$ in vacuum. The amplitude of the magnetic field of the wave is close to (Take, $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 \frac{	ext{Nm}^2}{	ext{C}^2}$ and speed of light $= 3 	imes 10^8 	ext{ ms}^{-1}$)

Options:

(A) 190 nT

(B) 160 nT

(C) 180 nT

(D) 150 nT

Sigma Percentile · Accepted Answer

### The Anatomy of an Electromagnetic Wave

Imagine an electromagnetic wave traveling through the vastness of space. It is a beautiful, self-sustaining dance of energy, consisting of oscillating electric and magnetic fields that are perfectly in sync.

These fields don't just wave around; they carry energy. A fundamental principle of electromagnetism is that this total energy is shared **equally** between the electric field and the magnetic field.

When a problem gives you the "energy density" of the wave, it is referring to this total combined energy per unit volume.

### The Energy Density Connection

In this problem, we are given the total energy density $u = 1.02 	imes 10^{-8} 	ext{ J/m}^3$. We are also given the frequency of the wave.

But here is a classic exam trap! The frequency is completely irrelevant to finding the amplitude of the magnetic field. It is a red herring designed to test your confidence in the core formulas.

The master equation that connects the total average energy density $u$ to the magnetic field amplitude $B_0$ is:

$$u = \frac{B_0^2}{2\mu_0}$$

This elegant formula accounts for both the electric and magnetic contributions, wrapping them up neatly in terms of just the magnetic field and the permeability of free space, $\mu_0$.

### Navigating the Calculation

Our goal is to find $B_0$. Let's rearrange our master equation to isolate the magnetic field amplitude:

$$B_0 = \sqrt{2\mu_0 u}$$

Now, we substitute the given values. We know the standard value for the permeability of free space is $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.

$$B_0 = \sqrt{2 	imes (4\pi 	imes 10^{-7}) 	imes (1.02 	imes 10^{-8})}$$

Let's group the numbers and the powers of 10 to keep things clean:

$$B_0 = \sqrt{8\pi 	imes 1.02 	imes 10^{-15}}$$

Multiplying $8 	imes 3.14 	imes 1.02$ gives us approximately $25.6$. So our expression becomes:

$$B_0 = \sqrt{25.6 	imes 10^{-15}}$$

### The Final Reveal

Taking the square root of an odd power of 10 can be messy. Let's shift the decimal point to create an even power of 10, which is much easier to work with:

$$B_0 = \sqrt{2.56 	imes 10^{-14}}$$

This is where the magic happens. You might recognize that $256$ is a perfect square ($16^2 = 256$). Therefore, the square root of $2.56$ is exactly $1.6$.

$$B_0 = 1.6 	imes 10^{-7} 	ext{ T}$$

Finally, we need to match our answer to the given options, which are in nano-Tesla (nT). A nano-Tesla is $10^{-9} 	ext{ T}$. By shifting the decimal point two places to the right, we get:

$$B_0 = 160 	imes 10^{-9} 	ext{ T} = \mathbf{160 	ext{ nT}}$$

And there we have it! By trusting our core formulas and carefully navigating the algebra, we've arrived at the perfect answer.

Select Answer:

Visualized Solution

Visualizing the EM Wave

Given Parameters

The Energy Density Formula

Rearranging for $B_{0}$

Substituting the Values

Simplifying the Expression

Final Calculation

Converting to Nano-Tesla

The Sigma Insight: Characteristics of Electromagnetic Waves

The Anatomy of an Electromagnetic Wave

The Energy Density Connection

Navigating the Calculation

The Final Reveal

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