JEE Main 2017 · 8 Apr · Shift 1 · Physics Q21

← JEE Main 2017 · 8 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 21 of 90

Magnetic field in a plane electromagnetic wave is given by
$\vec B=B_0\sin(kx+\omega t)\hat j\ \text{T}$
Expression for corresponding electric field will be :

Where c is speed of light.

\vec E=B_0c\sin(kx+\omega t)\hat k

V/m

\vec E=\dfrac{B_0}c\sin(kx+\omega t)\hat k

V/m

\vec E=-B_0c\sin(kx+\omega t)\hat k

V/m

\vec E=B_0c\sin(kx-\omega t)\hat k

V/m

Answer and worked solution

Official answer: A

The phase $kx+\omega t$ means the wave travels along $-\hat i$ . $\vec E\times\vec B$ must point along $-\hat i$ : with $\vec B$ along $\hat j$ , $\hat k\times\hat j=-\hat i$ , so $\vec E$ is along $+\hat k$ with magnitude $cB$ :
$\vec E=B_0c\sin(kx+\omega t)\hat k.$

Topics:Properties of EM waves

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 21