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NEET
Exam
NEET
Subject
Physics
Chapter
Electromagnetic Waves

NEET · Physics · Electromagnetic Waves

The electric field of a plane electromagnetic wave propagating in a...

The electric field of a plane electromagnetic wave propagating in a transparent non-magnetic medium is given by E⃗=40sin⁡(1.2×108t−0.6z)i^ V/m\vec{E} = 40 \sin(1.2 \times 10^8 t - 0.6 z) \hat{i} \text{ V/m}. The corresponding magnetic field equation will be:

Options

  1. A2×10−7sin⁡(1.2×108t−0.6z)j^ T2 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}
  2. B1.33×10−7sin⁡(1.2×108t−0.6z)j^ T1.33 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}
  3. C−2×10−7sin⁡(1.2×108t−0.6z)j^ T-2 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}
  4. D8×109sin⁡(1.2×108t−0.6z)j^ T8 \times 10^9 \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}

Correct answer

A. 2×10−7sin⁡(1.2×108t−0.6z)j^ T2 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}

Step-by-step solution

Given the electric field equation: E⃗=40sin⁡(1.2×108t−0.6z)i^ V/m\vec{E} = 40 \sin(1.2 \times 10^8 t - 0.6 z) \hat{i} \text{ V/m}
Comparing with the standard equation E⃗=E0sin⁡(ωt−kz)i^\vec{E} = E_0 \sin(\omega t - kz) \hat{i}, we get:
E0=40 V/mE_0 = 40 \text{ V/m}
ω=1.2×108 rad/s\omega = 1.2 \times 10^8 \text{ rad/s}
k=0.6 rad/mk = 0.6 \text{ rad/m}

The speed of the wave in the medium is:
v=ωk=1.2×1080.6=2×108 m/sv = \dfrac{\omega}{k} = \dfrac{1.2 \times 10^8}{0.6} = 2 \times 10^8 \text{ m/s}

The amplitude of the magnetic field is:
B0=E0v=402×108=2×10−7 TB_0 = \dfrac{E_0}{v} = \dfrac{40}{2 \times 10^8} = 2 \times 10^{-7} \text{ T}

The wave propagates in the +z+z direction (since the phase is ωt−kz\omega t - kz), so v^=k^\hat{v} = \hat{k}.
The direction of the magnetic field is given by E^×B^=v^\hat{E} \times \hat{B} = \hat{v}.
i^×B^=k^⇒B^=j^\hat{i} \times \hat{B} = \hat{k} \Rightarrow \hat{B} = \hat{j}

Therefore, the magnetic field equation is:
B⃗=2×10−7sin⁡(1.2×108t−0.6z)j^ T\vec{B} = 2 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}

Answer: 2×10−7sin⁡(1.2×108t−0.6z)j^ T2 \times 10^{-7} \sin(1.2 \times 10^8 t - 0.6 z) \hat{j} \text{ T}
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