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Electromagnetic Waves JEE 2025: Photon Question Solved

JEE Advanced 2025 Physics Electromagnetic Waves Energy density of electromagnetic waves

By Founder, JEEnius - IIT Kanpur Alumni · Oct 13, 2026 · 4 min read

Medium 2 min target

A cube of unit volume contains 35×107 photons of frequency 1015 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 α×10−9 T. Taking permeability of free space μ0=4π×10−7 Tm/A, Planck's constant h=6×10−34 Js and π=227, the value of α is ________.

Show answerAnswer

22.98

Explanation

Total photon energy is the number of photons times energy per photon.

E=Nhf

Here,

N=35×107

h=6×10−34 Js

f=1015 Hz

So,

E=35×107×6×10−34×1015

For an electromagnetic wave, the average total energy density in terms of magnetic field amplitude B0 is

uavg=B022μ0

The cube has unit volume, so the total energy equals the average energy density.

E=B022μ0

Therefore,

B02=2μ0E

Substitute values:

B02=2×4×227×10−7×35×107×6×10−34×1015

Combining powers of 10:

10−7×107×10−34×1015=10−19

Combining numerical factors:

2×4×227×35×6=5280

Hence,

B02=5280×10−19

B02=528×10−18

Taking square root:

B0=528×10−9 T

B0=22.978×10−9 T

Given,

B0=α×10−9 T

So,

α=22.978

Rounded to two decimal places:

α=22.98

Final answer: 22.98

Physics artwork for the article: Electromagnetic Waves JEE 2025: Photon Question Solved

What is the answer to the electromagnetic waves JEE 2025 photon question?

The required coefficient is 22.98 in the electromagnetic waves JEE 2025 photon question, found by equating photon energy per unit volume to average total energy density. This is JEE Advanced 2025, Paper 1, Physics, not JEE Main. It is a numerical-answer question, rated medium on the question bank’s scale, with a suggested solving target of 120 seconds.

A cube contains photons with the following count and frequency:

N=35×107,f=1015 Hz,V=1 m3.

Their combined energy represents the average electromagnetic energy in that volume. Find the coefficient in the magnetic-field amplitude:

B0=α×10−9 T.

Use the supplied constants:

h=6×10−34 Js,μ0=4π×10−7 Tm/A,π=227.

How do you calculate the total energy carried by the photons?

Multiply the photon count by the energy per photon, as in the official solution. Define the symbols and write:

E=total photon energy,N=number of photons.

E=Nhf. First calculate the energy per photon. Then multiply by the photon count, keeping numerical factors and powers of ten separate:

hf=(6×10−34)(1015)=6×10−19 J.
E=Nhf=(35×107)(6×10−34)(1015).
35×6=210,107−34+15=10−12.
E=210×10−12 J=2.10×10−10 J.

Divide this total energy by the volume to obtain the average energy density. The volume is one cubic metre:

uavg=EV=2.10×10−10 J1 m3=2.10×10−10 J/m3.

Energy and energy density have equal numerical values here, not identical units. Unit volume simplifies the arithmetic; it does not remove the physical distinction.

Which energy-density formula uses the peak magnetic field?

The official relation connects average total energy density to the peak magnetic-field amplitude:

uavg=B022μ0.

The photon energy includes both electric and magnetic energy, not the magnetic contribution alone. At any instant, the magnetic energy density is:

uB=B22μ0.

The electric and magnetic contributions are equal in the wave. Their instantaneous total is therefore:

utotal=uE+uB=B2μ0.

For a sinusoidal wave, the cycle average of the squared field is half its peak value. Applying that average gives:

⟨B2⟩=B022,⟨utotal⟩=B022μ0.

Adding the electric contribution doubles the magnetic energy; time averaging then introduces a factor of one-half. These are separate steps, even though the final formula resembles the instantaneous magnetic-energy formula.

Equate the average total density to photon energy per volume, then rearrange:

EV=B022μ0⇒B02=2μ0EV.

For the specified unit volume, the official numerical substitution is:

B02=2μ0E(unit-volume SI substitution).

Division by one cubic metre remains understood. This shorthand is not a general equation for other volumes.

How does the calculation give the coefficient 22.98?

The squared amplitude reduces to 528 multiplied by ten to the power negative eighteen. Taking its square root gives the required coefficient, rounded to 22.98.

Using the unit volume, substitute the constants exactly as supplied:

B02=2×4×227×10−7×35×107×6×10−34×1015 T2.

Combine the exponents first, then the numerical factors. Keeping these calculations separate avoids mixing coefficient arithmetic with powers of ten: −7+7−34+15=−19.

2×4×227×35×6=5280.
B02=5280×10−19 T2=528×10−18 T2.

Use the even exponent before taking the square root. This avoids a half-integer exponent:

B0=528×10−9 T≈22.978×10−9 T.

Compare directly with the requested form. The power of ten is already included in the question:

B0=α×10−9 T⇒α=528≈22.98.

Enter 22.98, because the question asks for the coefficient, not the field value:

Required entry: 22.98,not 22.98×10−9.

Retain the supplied approximations for Planck’s constant and pi. Replacing them with more precise constants changes the numerical problem being answered.

Why would using peak energy instead of average energy give 16.25?

The result 16.25 comes from treating peak instantaneous total energy density as the cycle average. This is a numerical-answer question with no supplied multiple-choice options: 16.25 is an erroneous result to diagnose, not a listed option.

The incorrect starting equation is:

uavg=B02μ0(incorrect for peak amplitude).

It omits the time-average factor. Compare the resulting squared field with the correct expression:

Bwrong2=μ0uavg,B02=2μ0uavg.
αwrong=5282=264≈16.25.

The incorrect coefficient is smaller by the following factor:

αwrongα=12.

It actually gives the RMS magnetic field, not the requested peak amplitude:

Brms=B02≈16.25×10−9 T.

Before substituting, label the energy as instantaneous or averaged and the field as peak or RMS. Select the formula from those labels, not merely because it contains magnetic field.

How do photon count and volume change the answer?

At fixed frequency, peak field grows with the square root of photon count and falls with the square root of volume. These three questions are original practice variations, not additional verified PYQs. Use the same constants and frequency unless explicitly changed.

What is the coefficient if the photon count becomes four times larger?

The coefficient doubles to 45.96. The new photon count and unchanged volume are:

N′=140×107,V′=1 m3.

Four times the photon count gives four times the energy density. Field amplitude follows its square root, so it doubles rather than quadruples:

B0∝N,α′=2528≈45.96.

What is the coefficient if the original photons occupy four cubic metres?

The coefficient halves to 11.49. Spreading the same photon energy through four times the volume reduces energy density to one-quarter:

V′=4 m3,B0∝1V.
α′=5282≈11.49.

What are the separately averaged electric and magnetic energy densities?

Each is half the original average total energy density. The photon energy is shared equally between the two contributions, not assigned in full to each:

⟨uE⟩=⟨uB⟩=uavg2=1.05×10−10 J/m3.

For another variation, use the same photon-energy method. Keep volume explicit until substitution:

B0=2μ0NhfV.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with How to Study Circles JEE: A Six-Step Plan.

Frequently asked questions

What is the answer to the electromagnetic waves JEE 2025 photon question?

The required coefficient is 22.98 in JEE Advanced 2025, Paper 1, Physics. Using the supplied constants gives B₀ = √528 × 10⁻⁹ T, so enter 22.98 rather than 22.98 × 10⁻⁹.

Which energy-density formula should I use for peak magnetic field?

For a sinusoidal electromagnetic wave, average total energy density is u_avg = B₀²/(2μ₀), where B₀ is the peak magnetic-field amplitude. Equating this to photon energy per unit volume, Nhf/V, gives B₀ = √(2μ₀Nhf/V). The total includes both electric and magnetic contributions.

Why am I getting 16.25 instead of 22.98?

Using u_avg = B₀²/μ₀ incorrectly treats peak instantaneous total energy density as the cycle average. This gives a coefficient of about 16.25, which corresponds to the RMS magnetic field rather than the requested peak amplitude. Multiply the RMS field by √2 to recover the peak field.

How does magnetic-field amplitude depend on photon count and volume?

At fixed frequency, peak magnetic-field amplitude is proportional to √(N/V). Quadrupling the photon count at unchanged volume doubles the coefficient from 22.98 to 45.96. Quadrupling the volume with the original photon count halves it to 11.49.

electromagnetic wavesenergy densityjee advanced 2025peak and rmsphoton energy

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