A cube of unit volume contains 35 x 10^7 photons of frequency 10^15 Hz. If the energy of all the pho

Q: Why do we use $u = \frac{B_0^2}{2\mu_0}$ instead of $u = \frac{B_0^2}{4\mu_0}$?

The formula $\frac{B_0^2}{4\mu_0}$ represents only the *magnetic* part of the average energy density. An EM wave consists of both electric and magnetic fields, which contribute equally. The total average energy density is the sum of both, which simplifies to $\frac{B_0^2}{2\mu_0}$.

Q: Can we solve this using the electric field amplitude $E_0$ first?

Absolutely! You could equate the total energy to $\frac{1}{2}\epsilon_0 E_0^2$ to find $E_0$, and then use the relation $E_0 = c B_0$ to find $B_0$. However, using the magnetic energy density directly saves a step and reduces calculation errors.

Q: Why is the volume taken as $1 \text{ m}^3$?

The problem explicitly states 'A cube of unit volume'. In physics, 'unit volume' means a volume of 1 in the standard SI units, which is $1 \text{ m}^3$.

Q: How do we quickly estimate the square root of 528?

Memorizing squares up to 25 is a lifesaver in JEE. We know $23^2 = 529$. Since 528 is extremely close to 529, its square root will be just slightly less than 23. The value 22.98 is a very safe and accurate estimate.

Q: What if the photons were not in a vacuum?

If the medium wasn't a vacuum, the speed of light would change, and we would use the permeability $\mu$ and permittivity $\epsilon$ of that specific medium instead of $\mu_0$ and $\epsilon_0$. The fundamental energy equivalence would still hold.

Question

A cube of unit volume contains $35 	imes 10^7$ photons of frequency $10^{15} 	ext{ Hz}$. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $\alpha 	imes 10^{-9} 	ext{ T}$. Taking permeability of free space $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ Tm/A}$, Planck's constant $h = 6 	imes 10^{-34} 	ext{ Js}$ and $\pi = \frac{22}{7}$, the value of $\alpha$ is____

Sigma Percentile · Accepted Answer

The problem we are tackling today is a beautiful bridge between two monumental pillars of physics: Quantum Mechanics and Classical Electromagnetism. On one side, we have photons—discrete, indivisible packets of energy. On the other side, we have the continuous, undulating waves of electric and magnetic fields.

This question asks us to look at a single volume of space and equate the energy described by these two entirely different paradigms. Let's dive into this fascinating duality!

### The Quantum Perspective: Energy of Photons

Imagine a cube with a volume of exactly $1 	ext{ m}^3$. Inside this cube, a staggering number of photons are buzzing around. We are given that the number of photons is $N = 35 	imes 10^7$ and each photon has a frequency of $f = 10^{15} 	ext{ Hz}$.

According to Max Planck and Albert Einstein, the energy of a single photon is directly proportional to its frequency. The formula is elegantly simple:
$$E = hf$$
where $h$ is Planck's constant ($6 	imes 10^{-34} 	ext{ Js}$).

Since we have $N$ photons, the total energy $U$ contained within our cubic meter is simply the energy of one photon multiplied by the total number of photons:
$$U = N \cdot hf$$

Let's substitute the values and calculate this total energy:
$$U = (35 	imes 10^7) 	imes (6 	imes 10^{-34}) 	imes (10^{15})$$

Grouping the numbers and the powers of ten makes the calculation a breeze:
$$U = (35 	imes 6) 	imes 10^{7 - 34 + 15}$$
$$U = 210 	imes 10^{-12} 	ext{ J}$$

To keep things in standard scientific notation, we can rewrite this as:
$$U = 2.1 	imes 10^{-10} 	ext{ J}$$

This is the total energy residing in our cube, calculated purely from a quantum mechanical standpoint.

### The Classical Perspective: Energy Density of EM Waves

Now, let's switch gears. The problem states that this exact same energy can be viewed as the average energy of an electromagnetic wave occupying the same volume.

In classical electromagnetism, an EM wave carries energy through its oscillating electric and magnetic fields. The energy is spread out over space, which is why we talk about **energy density** (energy per unit volume), denoted by $u$.

The total average energy density of an EM wave is the sum of the energy densities of the electric field and the magnetic field. Remarkably, in an EM wave, the electric and magnetic fields contribute equally to the total average energy.

Expressed in terms of the magnetic field amplitude $B_0$, the total average energy density is:
$$u = \frac{B_0^2}{2\mu_0}$$
where $\mu_0$ is the permeability of free space ($4\pi 	imes 10^{-7} 	ext{ Tm/A}$).

### Bridging the Two Worlds

We know the total energy $U$, and we know the volume $V = 1 	ext{ m}^3$. Since energy density is energy divided by volume ($u = \frac{U}{V}$), the total energy is simply the energy density multiplied by the volume:
$$U = u 	imes V$$

Substituting our classical expression for $u$:
$$U = \frac{B_0^2}{2\mu_0} 	imes V$$

This is the master equation where the quantum world meets the classical world. Let's plug in the quantum energy we calculated earlier:
$$2.1 	imes 10^{-10} = \frac{B_0^2}{2 	imes (4\pi 	imes 10^{-7})} 	imes 1$$

### The Final Calculation

Our goal is to find the amplitude of the magnetic field, $B_0$. Let's isolate $B_0^2$ by multiplying both sides by the denominator:
$$B_0^2 = 2.1 	imes 10^{-10} 	imes 8\pi 	imes 10^{-7}$$

The problem kindly provides the approximation $\pi = \frac{22}{7}$. This is a massive hint that things are going to cancel out beautifully. Let's substitute it in:
$$B_0^2 = 2.1 	imes 10^{-10} 	imes 8 	imes \frac{22}{7} 	imes 10^{-7}$$

Look at the $2.1$ and the $7$. This is where you need to avoid silly mistakes. Dividing $2.1$ by $7$ gives exactly $0.3$:
$$B_0^2 = 0.3 	imes 8 	imes 22 	imes 10^{-17}$$

Now, let's multiply the numbers:
$$0.3 	imes 8 = 2.4$$
$$2.4 	imes 22 = 52.8$$

So, we have:
$$B_0^2 = 52.8 	imes 10^{-17}$$

To make taking the square root easier, let's shift the decimal point to make the power of ten an even number:
$$B_0^2 = 528 	imes 10^{-18}$$

Finally, we take the square root of both sides:
$$B_0 = \sqrt{528} 	imes 10^{-9} 	ext{ T}$$

How do we find the square root of $528$ without a calculator? This is where your mental math toolkit comes in handy. You should know that $23^2 = 529$. Since $528$ is incredibly close to $529$, its square root will be just a hair under $23$.

A highly accurate estimate is $22.98$. Therefore:
$$B_0 \approx 22.98 	imes 10^{-9} 	ext{ T}$$

The problem asks for the value of $\alpha$ where $B_0 = \alpha 	imes 10^{-9} 	ext{ T}$. Comparing our result, we find:
$$\mathbf{\alpha = 22.98}$$

And there we have it! By seamlessly transitioning between the particle nature and the wave nature of light, we've arrived at our final answer.

Enter Numerical Value:

Visualized Solution

$V = 1 m^{3}, N = 35 \times 1 0^{7}, f = 1 0^{15} Hz$

$U = N \cdot h f$

$U = (35 \times 1 0^{7}) \times (6 \times 1 0^{- 34}) \times (1 0^{15})$

$U = 2.1 \times 1 0^{- 10} J$

$u = \frac{B _{0}^{2}}{2 μ _{0}}$

$U = u \times V$

$B_{0}^{2} = 2.1 \times 1 0^{- 10} \times 2 μ_{0}$

$B_{0}^{2} = 2.1 \times 1 0^{- 10} \times 8 \times \frac{22}{7} \times 1 0^{- 7}$

$B_{0}^{2} = 528 \times 1 0^{- 18}$

$B_{0} = 22.98 \times 1 0^{- 9} T$

The Sigma Insight: Characteristics of Electromagnetic Waves

The Quantum Perspective

The Classical Perspective

Bridging the Two Worlds

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=1 m3,N=35×107,f=1015 Hz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=N⋅hf

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=(35×107)×(6×10−34)×(1015)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=2.1×10−10 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=2μ0​B02​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=u×V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B02​=2.1×10−10×2μ0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B02​=2.1×10−10×8×722​×10−7

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B02​=528×10−18

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B0​=22.98×10−9 T