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Electromagnetic Waves question

2022 · 30 Jun · Shift 1 · Q54
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  5. /2022 · 30 Jun · Shift 1 · Q54

Electromagnetic Waves question

2022 · 30 Jun · Shift 1 · Q54

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5 π×\pi\timesπ× 103x −-− 3 π×\pi\timesπ× 1011t) Vm −-− 1 Then, the value of magnetic field amplitude will be : (Given : speed of light in Vacuum c = 3 ×\times× 108 ms −-− 1)
  1. A
    1 ×\times× 10 −-− 6 T
  2. B
    5 ×\times× 10 −-− 6 T
  3. C
    18 ×\times× 109 T
  4. D
    21 ×\times× 109 T
View written solutionFree

Correct answer: A

  1. Given electric field equation

    The plane electromagnetic wave is: Ez=300sin⁡(5π×103x−3π×1011t) V m−1E_z = 300\sin\left(5\pi\times 10^3 x - 3\pi\times 10^{11} t\right)\, \text{V m}^{-1}Ez​=300sin(5π×103x−3π×1011t)V m−1

    Comparing with the standard form: E=E0sin⁡(kx−ωt)E = E_0 \sin(kx-\omega t)E=E0​sin(kx−ωt)

    we get the electric field amplitude: E0=300 V m−1E_0 = 300\, \text{V m}^{-1}E0​=300V m−1

  2. Relation between electric and magnetic field amplitudes

    For an electromagnetic wave in vacuum: E0=cB0E_0 = cB_0E0​=cB0​

    Hence, B0=E0cB_0 = \frac{E_0}{c}B0​=cE0​​

  3. Substitute the given values

    B0=3003×108B_0 = \frac{300}{3\times 10^8}B0​=3×108300​

    B0=100×10−8B_0 = 100\times 10^{-8}B0​=100×10−8

    B0=10−6 TB_0 = 10^{-6}\, \text{T}B0​=10−6T

  4. Final answer

    B0=1×10−6 T\boxed{B_0 = 1\times 10^{-6}\, \text{T}}B0​=1×10−6T​

  5. Option check

    • A: 1×10−6 T1\times 10^{-6}\,\text{T}1×10−6T ✅
    • B: 5×10−6 T5\times 10^{-6}\,\text{T}5×10−6T ❌
    • C: 18×109 T18\times 10^{9}\,\text{T}18×109T ❌
    • D: 21×109 T21\times 10^{9}\,\text{T}21×109T ❌

Therefore, the correct option is A.

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