The equation that represents magnetic field of a plane electromagnetic wave which is propagating along…

The equation that represents magnetic field of a plane electromagnetic wave which is propagating along $x$-direction with wavelength $10 \mathrm{~mm}$ and maximum electric field $60 \mathrm{Vm}^{-1}$ in $y$-direction is (where, $c=$ speed of light)
  1. $\left(6 \times 10^{-7}\right) \sin [0.2 \pi(c t-x)] \hat{k} T$
  2. $\left(2 \times 10^{-7}\right) \sin [200 \pi(c t-x)] \hat{k} T$
  3. $\left(2 \times 10^{-7}\right) \sin [200 \pi(c t-x)] \hat{i} \mathrm{~T}$
  4. $\left(6 \times 10^{-7}\right) \sin [02 \pi(c t-x)] \hat{i} T$

Solution

Equation of magnetic field of an electromagnetic wave is given by $B=B_0 \sin (\omega t-k x) \hat{k}$ $=B_0 \sin \left(2 \pi f t-\frac{2 \pi}{\lambda} \cdot x\right) \hat{k} \quad\left[\because \omega=2 \pi f\right.$ and $\left.k=\frac{2 \pi}{\lambda}\right]$ $=B_0 \sin \frac{2 \pi}{\lambda}[f \lambda t-x] \hat{k}$ $B=B_0 \sin \frac{2 \pi}{\lambda}[c t-x] \hat{k}$ ...(i) Also direction of $\mathbf{B}$ vector is perpendicular to $\mathbf{E}$ and also to direction of propagation And $B_0=\frac{E_0}{c}$ Here, $E_0=60 \mathrm{Vm}^{-1}$ $\therefore \quad B_0=\frac{60^{}}{3 \times 10^8}=2 \times 10^{-7}$ Also $k=\frac{2 \pi}{\lambda}$ Here $k=200 \pi=\frac{2 \pi}{\lambda}$ $\Rightarrow \quad \lambda=\frac{1}{100} \mathrm{~m}=\frac{100}{100} \mathrm{~cm}=10 \mathrm{~mm}=10^{-2} \mathrm{~m}$ Putting the values is eq. (i), we get $B=2 \times 10^{-7} \sin \left[\frac{2 \pi}{10^{-2}}(c t-x)\right] \hat{k}$ $=2 \times 10^{-7} \sin 200 \pi(t-x) \hat{k}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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