Electromagnetic Waves

JEE Main level, single correct, medium

7 of 165

Question 7 of 165

A magnetic field vector in an electromagnetic wave is represented by B⃗=B0sin⁡(2πvt−2πxλ)j^\vec{B}=B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{j}. Its associated electric field vector is ______.

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Answer: (A) E⃗=−vλB0sin⁡(2πvt−2πxλ)k^\vec{E}=-v\lambda B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{k}

The direction of propagation of the electromagnetic wave is determined by the phase term 2πvt−2πxλ2\pi vt - \dfrac{2\pi x}{\lambda}, which indicates that the wave is travelling along the +x+x direction (i^\hat{i}).

The direction of the magnetic field B⃗\vec{B} is given as j^\hat{j}.

In an electromagnetic wave, the direction of propagation is given by the direction of the cross product E⃗×B⃗\vec{E} \times \vec{B}.

Let the direction of the electric field E⃗\vec{E} be n^\hat{n}.

n^×j^=i^\hat{n} \times \hat{j} = \hat{i}

Since (−k^)×j^=i^(-\hat{k}) \times \hat{j} = \hat{i}, the direction of the electric field must be −k^-\hat{k}.

The amplitude of the electric field is related to the magnetic field by E0=cB0E_0 = c B_0, where cc is the speed of the wave.

The speed of the wave is c=vλc = v \lambda, where vv is the frequency.

E0=vλB0E_0 = v \lambda B_0

Therefore, the electric field vector is E⃗=−vλB0sin⁡(2πvt−2πxλ)k^\vec{E} = -v \lambda B_0 \sin\left(2\pi vt - \dfrac{2\pi x}{\lambda}\right) \hat{k}.

Answer: E⃗=−vλB0sin⁡(2πvt−2πxλ)k^\vec{E}=-v\lambda B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{k}

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