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PhysicsMediumClass 12

Energy and Intensity of EM waves

EM Waves

8

JEE Qs

8%

Hard

75

min

Master the distinction between instantaneous, peak, and RMS values for electric and magnetic fields, and understand their impact on energy density and intensity calculations.

🧮 Key Formulas

u_e = (1/2)ε₀E²
u_m = (1/2)(B²/μ₀)
u = u_e + u_m = ε₀E² = B²/μ₀ (instantaneous total energy density)
<u_avg> = (1/2)ε₀E₀² = (1/2)(B₀²/μ₀) = ε₀E_rms² = B_rms²/μ₀ (average total energy density)
S = (1/μ₀)(E x B) (Poynting vector)
|S| = EB/μ₀
I = <|S|> = (1/2)ε₀cE₀² = (1/2)(c/μ₀)B₀² = ε₀cE_rms² = (c/μ₀)B_rms² (average intensity)
E₀ = cB₀
E_rms = E₀/sqrt(2)
B_rms = B₀/sqrt(2)
I = P/A (Intensity as Power per Area)

✅ Key Points for JEE

  • 1In an electromagnetic wave, the energy is equally distributed between the electric field and the magnetic field.
  • 2The Poynting vector (S) represents the instantaneous power per unit area and its direction indicates the direction of energy propagation of the EM wave.
  • 3Intensity (I) is the average power per unit area carried by the EM wave over one complete cycle and is proportional to the square of the peak (or RMS) electric or magnetic field amplitude.
  • 4EM waves carry both energy and momentum. The intensity is directly related to the average energy density and the speed of light (I = <u_avg> * c).
  • 5Be precise with using peak values (E₀, B₀) versus RMS values (E_rms, B_rms) in intensity and energy density formulas, remembering the factor of 1/2 for peak values and no factor for RMS values in average expressions.

⚠️ Common Mistakes

  • Confusing instantaneous field values with peak values (E₀, B₀) or RMS values (E_rms, B_rms), especially in formulas involving average intensity or energy density, often leading to missing or incorrectly applying factors of 1/2 or sqrt(2).
  • Incorrectly relating the electric and magnetic field magnitudes (E=cB) or the Poynting vector magnitude (S=EB/μ₀) when the fields are not perpendicular or for situations outside of a plane EM wave.
  • Errors in unit conversion or using incorrect values for fundamental constants (ε₀, μ₀, c), which are critical for numerical problems.

📝 Practice Questions

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Q39.Arrange the following in the ascending order of wavelength (λ) : (A) Microwaves (λ1) (B) Ultraviolet rays (λ2) (C) Infrared rays (λ3) (D) X-rays (λ4) Choose the most appropriate answer from the options given below : (1) λ4 < λ3 < λ2 < λ1 (2) λ3 < λ4 < λ2 < λ1 (3) λ4 < λ3 < λ1 < λ2 (4) λ4 < λ2 < λ3 < λ1

2025·MCQEasy

Q38.The magnetic field of an E.M. wave is given by B = ( √32 ^i + 12 ^j)30 corresponding electric field in S.I. units is : → → (1) 1 (2) 3 1 E = cos [ω (t −zc )] ( 2^i −√32 ^j)30c sin [ω (t −zc )] E = ( 4^i + 4 ^j)30c → → √3 (3) (4) E = + E = sin [ω (t + zc )] sin [ω (t + zc )] ( 12^i 2 ^j)30c ( √32 ^i −12 ^j)30c

2025·MCQMedium

Q37.The electric field of an electromagnetic wave in free space is → E = 57 cos [7.5 × 106t −5 × 10−3(3x + 4y)](4^i −3^j)N/C . The associated magnetic field in Tesla is ⇀ 57 (1) B= cos [7.5 × 106t −5 × 10−3(3x + 4y)](^k) 3×108 → (2) 57 B = − cos [7.5 × 106t −5 × 10−3(3x + 4y)](^k) 3×108 → (3) 57 B = − cos [7.5 × 106t −5 × 10−3(3x + 4y)](5^k) 3×108 → (4) 57 B = cos [7.5 × 106t −5 × 10−3(3x + 4y)](5^k) 3×108

2025·MCQHard

Q39.The equation of a transverse wave travelling along a string is y(x, t) = 4.0 sin [20 × 10−3x + 600t]mm , where x is in mm and t is in second. The velocity of the wave is : 2025 (23 Jan Shift 2) JEE Main Previous Year Paper (1) −60 m/s (2) −30 m/s (3) +30 m/s (4) +60 m/s

2025·MCQEasy

Q38.Due to presence of an em-wave whose electric component is given by E = 100 sin(ωt −kx)NC−1 , a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as (1) 400 sin(ωt −kx)NC−1 (2) 200 sin(ωt −kx)NC−1 (3) 50 sin(ωt −kx)NC−1 (4) 25 sin(ωt −kx)NC−1

2025·MCQMedium

Q30.Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Electromagnetic waves carry energy but not momentum. Reason ( R): Mass of a photon is zero. In the light of the above statements, choose the most appropriate answer from the options given below : (1) Both (A) and (R) are true and (R) is the correct (2) Both (A) and (R) are true but (R) is not the explanation of (A) correct explanation of (A) (3) (A) is false but (R) is true (4) (A) is true but ( R) is false

2025·Assertion ReasoningEasy

NCERT Chapters

  • Class 12 Physics Ch 8: Electromagnetic Waves

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