Electromagnetic Waves

JEE Main level, single correct, medium

111 of 165

Question 111 of 165

The electric field of a plane polarized electromagnetic wave in free space at time t=0t=0 is given by the expression E⃗(x,y)=10j^cos(6x+8z)\vec{E}\left(x, y\right)=10\hat{j}cos\left(6x+8z\right). The magnetic field B⃗(x,z,t)\vec{B}\left(x, z, t\right) is given by (cc is the velocity of light.)

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Answer: (D) 1c(6k^−8i^)cos(6x+8z−10ct)\dfrac{1}{c}\left(6\hat{k}-8\hat{i}\right)cos\left(6x+8z-10 ct\right)

E=E0cos⁡(ωt−K⃗⋅r⃗)E={E}_{0}\cos \left(\omega t-\vec{K}\cdot \vec{r}\right)

∴∴ At any time tt,

K⃗=6i^+8k^\vec{K}=6\hat{i}+8\hat{k}

B0=E0c=10c{B}_{0}=\dfrac{{E}_{0}}{c}=\dfrac{10}{c}

λ=2πk=2π10=π5\lambda =\dfrac{2\pi }{k}=\dfrac{2\pi }{10}=\dfrac{\pi }{5}

Also, E⃗×B⃗=c⃗\vec{E}\times \vec{B}=\vec{c}

∴ω=2πf=2πcλ∴\omega =2\pi f=\dfrac{2\pi c}{\lambda }

∣i^j^k^010BxByBz∣=6i^+8k^10\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ {B}_{x} & {B}_{y} & {B}_{z}\end{matrix}\right|=\dfrac{6\hat{i}+8\hat{k}}{10}

ω=2π×c(π5)=10c\omega =\dfrac{2\pi \times c}{\left(\dfrac{\pi }{5}\right)}=10c

∴i^[Bz−0]+j^(0)+k^(−Bx)=0.6i^−0.8k^∴\hat{i}\left[{B}_{z}-0\right]+\hat{j}\left(0\right)+\hat{k}\left(-{B}_{x}\right)=0.6\hat{i}-0.8\hat{k}

Comparing the coefficients on both sides,

∴Bz=0.6;Bx=−0.8∴{B}_{z}=0.6;{B}_{x}=-0.8

∣B0⃗∣=10c(−0.8i^+0.6k^)=−8i^+6k^c\left|\vec{{B}_{0}}\right|=\dfrac{10}{c}\left(-0.8\hat{i}+0.6\hat{k}\right)=\dfrac{-8\hat{i}+6\hat{k}}{c}

So the magnetic field,B⃗=1c(−8i^+6k^)cos⁡⁡(6x+8z−10ct)tesla\vec{B}=\dfrac{1}{c}\left(-8\hat{i}+6\hat{k}\right)\cos ⁡\left(6x+8z-10ct\right)tesla

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