A laser beam has intensity of 4.0 x 1014 W/m2. The amplitude of magnetic field associated with beam is ___

Year	2026	2025	2024	Total
Questions	10	12	9	31

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

I = (1)/(2) ε₀ E₀² c

B₀ = (E₀)/(c) = (1)/(c) \sqrt((2I)/(ε₀ c))

Core Logic

Expressing electric field amplitude $E₀$ $E_0$ in terms of intensity $I$ $I$:

E₀ = \sqrt((2I)/(ε₀ c))

Relating magnetic field amplitude $B₀$ $B_0$ to $E₀$ $E_0$ via $B₀ = (E₀)/(c)$ $B_0 = \frac{E_0}{c}$:

B₀ = (1)/(c)\sqrt((2I)/(ε₀ c)) = (1)/(3 \times 10⁸) 2 \times 4.0 \times 10¹⁴8.85 \times 10⁻¹² \times 3 \times 10⁸

Simplifying terms under the radical:

B₀ = (1)/(3 \times 10⁸) 8 \times 10¹⁴2.655 \times 10⁻³ = (1)/(3 \times 10⁸) 3.013 \times 10¹⁷ = (10)/(3) \sqrt((8)/(8.85 \times 3)) \approx 1.83 ~T

Step 1: Final Conclusion

The amplitude of the magnetic field is $1.83 ~T$ $1.83 \mathrm{~T}$.

Pattern Recognition

EM Wave Intensity: $I = (1)/(2) c (B₀²)/(μ₀) = (1)/(2) ε₀ E₀² c$ . Direct sub: $B₀ = \sqrt((2 μ₀ I)/(c))$ $B_0 = \sqrt{\frac{2 \mu_0 I}{c}}$ or $B₀ = (1)/(c) \sqrt((2I)/(ε₀ c)) \approx 1.83~T$ .

Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10¹⁰ ~Hz

and an amplitude of

50 ~Vm⁻¹

. The total average energy density of the electromagnetic field of the wave is : [Use

ε₀ = 8.85 \times 10⁻¹² C² / Nm²

]

A. $1.106 \times 10⁻⁸ Jm⁻³$
B. $4.425 \times 10⁻⁸ ~Jm⁻³$
C. $2.212 \times 10⁻⁸ ~Jm⁻³$
D. $2.212 \times 10⁻¹⁰ ~Jm⁻³$

Solution

Related Formula

Utotal average = (1)/(2)ε₀ E₀²

Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where $E₀$ $E_0$ is the amplitude of the electric field.

Step 2: Substitution

Given: $E₀ = 50 V/m$ $E_0 = 50 \mathrm{\, V/m}$ $ε₀ = 8.85 \times 10⁻¹² C²/(N\cdot m²)$

Uavg = (1)/(2) \times (8.85 \times 10⁻¹²) \times (50)²

Uavg = (1)/(2) \times 8.85 \times 10⁻¹² \times 2500

Uavg = 1.10625 \times 10⁻⁸ J/m³

Chapter Mix

Class 12 Physics: Electromagnetic Waves

A laser beam has intensity of $4.0 \times 10¹⁴$ $4.0 \times 10^{14}$ W/m $²$ $^2$. The amplitude of magnetic field associated with beam is ____ T. (Take $ε₀ = 8.85 \times 10⁻¹²$ $\epsilon_{0} = 8.85 \times 10^{-12}$ C $²$ $^2$/Nm $²$ $^2$ and c = 3 × 10 $⁸$ $^8$ m/s)

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

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A laser beam has intensity of 4.0 × 10¹⁴$4.0 \times 10^{14}$ W/m²$^2$. The amplitude of magnetic field associated with beam is ____ T. (Take ε₀ = 8.85 × 10⁻¹²$\epsilon_{0} = 8.85 \times 10^{-12}$ C²$^2$/Nm²$^2$ and c = 3 × 10⁸$^8$ m/s)

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 2: Substitution

Chapter Mix

More Electromagnetic Waves Questions — jee_main_2026_22_january_evening

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A laser beam has intensity of $4.0 \times 10¹⁴$ $4.0 \times 10^{14}$ W/m $²$ $^2$. The amplitude of magnetic field associated with beam is ____ T. (Take $ε₀ = 8.85 \times 10⁻¹²$ $\epsilon_{0} = 8.85 \times 10^{-12}$ C $²$ $^2$/Nm $²$ $^2$ and c = 3 × 10 $⁸$ $^8$ m/s)