The magnetic field in a plane electromagnetic wave is given as B=(3 × 10^ -7 ~T) (3 × 10^4 x+9 × 10^ 12 t) j The electric field of this wave is given as

Correct answer: B. 90 (3 × 10^4 x+9 × 10^ 12 t) k Vm^ -1 Given, Magnetic field in a plane EM wave, B=(3 × 10^ -7 ~T) (3 × 10^4 x+9 × 10^ 12 t) T Here, B_0=3 × 10^ -7 ~T Amplitude of electric field, aligned E_0 & =B_0 c & =3 × 10^ -7 × 3 × 10^8 aligned E_0=90 Vm^ -1 From the given equation of B , it is clear that E M wave is travelling along negative direction of X - axis i.e. If n be the unit vector along the direction of electric field, then n × j =- i Clearly, n = k Hence, E=E_0 (k x+ω t) k =90 (3 × 10^4 x+9 × 10^ 12 t) k Vm^ -1

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

The magnetic field in a plane electromagnetic wave is given as B=(3...

The magnetic field in a plane electromagnetic wave is given as

\mathbf{B}=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{j}

The electric field of this wave is given as

Options

A $90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{i} \mathrm{Vm}^{-1}$
B $90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$
C $45 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{i} \mathrm{Vm}^{-1}$
D $45 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$

Correct answer

B. $90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$

Step-by-step solution

Given, Magnetic field in a plane EM wave,

B=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \mathrm{T}

Here,

B_0=3 \times 10^{-7} \mathrm{~T}

Amplitude of electric field,

\begin{aligned} E_0 & =B_0 c \\ & =3 \times 10^{-7} \times 3 \times 10^8\end{aligned}

\Rightarrow \quad E_0=90 \mathrm{Vm}^{-1}

From the given equation of

, it is clear that

E M

wave is travelling along negative direction of

- axis i.e.

\therefore

If

\hat{\mathbf{n}}

be the unit vector along the direction of electric field, then

\hat{\mathbf{n}} \times \hat{j}=-\hat{i}

Clearly,

\hat{\mathbf{n}}=\hat{k}

Hence,

E=E_0 \sin (k x+\omega t) \hat{k}=90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$

Practice Electromagnetic Waves on Quantrex Academy →

More from Electromagnetic Waves

The electric field associated with an electromagnetic wave travelling in vacuum is given… JEE Advanced 2026 A monochromatic source of light operating at 15 kW emits 2.5× 10^ 22 photons/s. The… JEE Main 2026 For an electromagnetic wave propagating through vacuum, k , E and ω represent propagation… JEE Main 2026 A point light source emits E.M. waves in free space. A detector, placed at a distance of… JEE Main 2026 A displacement current of 4.0 A can be set up in the space between two parallel plates of… JEE Main 2026 A magnetic field vector in an electromagnetic wave is represented by B =B_0 (2π vt-(2π… JEE Main 2026 An electromagnetic wave travels in free space along the x -direction. At a particular… JEE Main 2026 An electromagnetic wave travelling in x -direction is described by field equation E_y =… JEE Main 2026 Full Electromagnetic Waves list All 99 Percentile Qs Bank for JEE Main PYQs