The electric field in a plane electromagnetic wave is given by E_z=60 (5 x+1.5 × 10^9 t) V / m . Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

Correct answer: A. B_y=2 × 10^ -7 (5 x+1.5 × 10^9 t) T aligned & v= ω k =(1.5 × 10^9)/(5)=3 × 10^8 ~m / s & B_0= F_0 v =(60)/(3 × 10^8)=20 × 10^ -8 =2 × 10^ -7 ~T aligned B_ y =2 × 10^ -7 (5 x+1.5 × 10^9 t) T

NEET 2025

Exam: NEET
Year: 2025
Subject: Physics
Chapter: Electromagnetic Waves

NEET 2025 · Physics · Electromagnetic Waves

The electric field in a plane electromagnetic wave is given by...

The electric field in a plane electromagnetic wave is given by

E_z=60 \cos \left(5 \mathrm{x}+1.5 \times 10^9 \mathrm{t}\right) \mathrm{V} / \mathrm{m} .

Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) : The electric field in a plane electromagnetic wave is given by E_z=60 (5 x+1.5 × 10^9 t) V / m . Then expression for

The electric field in a plane electromagnetic wave is given by E_z=60 (5 x+1.5 × 10^9 t) V / m . Then expression for

Options

A $B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$
B $B_x=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$
C $B_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) T$
D $B_y=60 \sin \left(5 x+1.5 \times 10^9 t\right) T$

Correct answer

A. $B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$

Step-by-step solution

\begin{aligned} & \mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{1.5 \times 10^9}{5}=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\ & \mathrm{B}_0=\frac{\mathrm{F}_0}{\mathrm{v}}=\frac{60}{3 \times 10^8}=20 \times 10^{-8}=2 \times 10^{-7} \mathrm{~T}\end{aligned}

\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \cos \left(5 \mathrm{x}+1.5 \times 10^9 \mathrm{t}\right) \mathrm{T}

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