Question

The magnetic field of a plane electromagnetic is given by [200π (y + ct)] . Where c = 3 × 108 m/s. The corresponding electric field is :-

Answer 1

E=6\times10^{10}\pi\cos(y+ct),\hat{x}

Answer 2

Solution:

We assume that the magnetic field is given by

$$\vec{B}=200\pi\cos(y+ct)\,\hat{z}$$

with $c=3\times10^8\,{\rm m/s}$. (Note: The form “$y+ct$” shows that the phase is constant when $y+ct=$ constant; such a wave propagates in the $-\hat{y}$ direction.)

For an electromagnetic wave in free space the fields satisfy:

$$\vec{B}=\frac{1}{c}\,\hat{n}\times\vec{E}\quad\Longrightarrow\quad \vec{E}=-c\,\hat{n}\times\vec{B}\,.$$

Since the phase is $y+ct$, the propagation direction unit vector is:

$$\hat{n}=-\hat{y}\,.$$

Now, calculate the cross product:

$$\hat{n}\times \vec{B} = (-\hat{y})\times\bigl[200\pi\cos(y+ct)\,\hat{z}\bigr] = -200\pi\cos(y+ct)\, (\hat{y}\times\hat{z})\,.$$

But, $\hat{y}\times\hat{z}=\hat{x}$, so

$$\hat{n}\times \vec{B} = -200\pi\cos(y+ct)\,\hat{x}\,.$$

Then,

$$\vec{E}=-c\,\Bigl[-200\pi\cos(y+ct)\,\hat{x}\Bigr]=200\pi\,c\,\cos(y+ct)\,\hat{x}\,.$$

Substitute $c=3\times10^8\,{\rm m/s}$:

$$\vec{E}=200\pi\,(3\times10^8)\cos(y+ct)\,\hat{x} = 6\times10^{10}\pi\cos(y+ct)\,\hat{x}\,.$$

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Explanation (Minimal):

1. Assume $\vec{B}=200\pi\cos(y+ct)\,\hat{z}$ so the wave travels in the $-y$ direction ($\hat{n}=-\hat{y}$).
2. Use $\vec{E}=-c\,\hat{n}\times\vec{B}$.
3. Compute $(-\hat{y})\times(\hat{z})=-\hat{x}$ so that $\vec{E}=200\pi\,c\,\cos(y+ct)\,\hat{x}$.
4. Substitute $c=3\times10^8$ to get the final answer.