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NEET
Exam
NEET
Subject
Physics
Chapter
Electromagnetic Waves

NEET · Physics · Electromagnetic Waves

The electric field of a plane electromagnetic wave in vacuum is...

The electric field of a plane electromagnetic wave in vacuum is given by E⃗=60sin⁡(1.5×107t−0.05z)i^ V/m\vec{E} = 60 \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{i} \text{ V/m}. The corresponding magnetic field vector equation will be:

Options

  1. A2×10−7sin⁡(1.5×107t−0.05z)(−j^) T2 \times 10^{-7} \sin \left(1.5 \times 10^7 t - 0.05 z\right) (-\hat{j}) \text{ T}
  2. B2×10−7sin⁡(1.5×107t−0.05z)j^ T2 \times 10^{-7} \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{j} \text{ T}
  3. C1.8×1010sin⁡(1.5×107t−0.05z)j^ T1.8 \times 10^{10} \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{j} \text{ T}
  4. D1.8×1010sin⁡(1.5×107t−0.05z)(−j^) T1.8 \times 10^{10} \sin \left(1.5 \times 10^7 t - 0.05 z\right) (-\hat{j}) \text{ T}

Correct answer

B. 2×10−7sin⁡(1.5×107t−0.05z)j^ T2 \times 10^{-7} \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{j} \text{ T}

Step-by-step solution

Given, E⃗=60sin⁡(1.5×107t−0.05z)i^ V/m\vec{E} = 60 \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{i} \text{ V/m}

Comparing with the standard equation E⃗=E0sin⁡(ωt−kz)i^\vec{E} = E_0 \sin(\omega t - k z) \hat{i}, we get:
E0=60 V/mE_0 = 60 \text{ V/m}
ω=1.5×107 rad/s\omega = 1.5 \times 10^7 \text{ rad/s}
k=0.05 rad/mk = 0.05 \text{ rad/m}

The speed of the wave is c=ωk=1.5×1070.05=3×108 m/sc = \dfrac{\omega}{k} = \dfrac{1.5 \times 10^7}{0.05} = 3 \times 10^8 \text{ m/s}.
The amplitude of the magnetic field is B0=E0c=603×108=2×10−7 TB_0 = \dfrac{E_0}{c} = \dfrac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}.

The argument (ωt−kz)(\omega t - k z) indicates that the wave is propagating in the +z+z direction, so the unit vector of propagation is k^\hat{k}.
The direction of propagation is given by E^×B^\hat{E} \times \hat{B}.
i^×B^=k^\hat{i} \times \hat{B} = \hat{k}
Since i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, the magnetic field must oscillate in the +j^+\hat{j} direction.

Therefore, the magnetic field equation is:
B⃗=2×10−7sin⁡(1.5×107t−0.05z)j^ T\vec{B} = 2 \times 10^{-7} \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{j} \text{ T}

Answer: 2×10−7sin⁡(1.5×107t−0.05z)j^ T2 \times 10^{-7} \sin \left(1.5 \times 10^7 t - 0.05 z\right) \hat{j} \text{ T}
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