NEET Physics Electromagnetic Waves Questions Solved

(b) Given root mean square value of electric field,

                $E_{rms}= 6 V/m$

We know that, peak value of electric field, 

   $E_o=\sqrt{2}{E}_{rms}$

$\Rightarrow E_o=\sqrt{2}\times 6 V/m$

Also, we know that, $c=\frac{E_o}{B_o}$

where, c= speed of light in vacuum 

$B_o= peak value of magnetic field$

$$\begin{gathered} \Rightarrow B_o=\frac{\sqrt{2}\times 6}{3\times {10}^8} \\ \Rightarrow B_o=\frac{8.48}{3}\times {10}^{-8} \\ \Rightarrow B_o=2.83\times {10}^{-8} T \end{gathered}$$

(c) An electric charge at rest has electric field in the region around it, but no magnetic field. A moving charge produces both the electric and magnetic fields. If a charge is moving with a constant velocity, the electric and magnetic fields will not change with time, hence no EM wave will be produced. But if the charge is moving with a non-zero acceleration, both the electric and magnetic field will change will space and time, it then produced EM wave. This shows that accelerated charge emits electromagnetic waves.

Given energy of EM waves is of the order of 15keV
i.e. E=hv=h x $\frac{c}{\lambda }$

=>λ=$\frac{h x c}{e}$

=6.624x10^-34 x 3x10¹⁸/15 x 10³ x 1.6x10^-19

=1.3248 x 10^-29/1.6x10^-19

=0.828x10^-10m

=0.828Å [∴1Å=10^-10m]

λ=0.828Å

Thus, this spectrum is a part of X-rays.

(a) It is electromagnetive wave,therefore the correct choice is (a).

Electromagnetic wave equation

          $E=E_0\cos \left(kz-\omega t\right)$                                 ...(i)

Speed of electromagnetic wave v=$\frac{\omega }{k}$

Given, equation

$E=40\cos \left(kz-6\times {10}^{8}t\right)$                                  ...(ii)

Comparing Eqs.(i) and (ii), we get

          $\omega =6\times {10}^8$

and    $E_0=40$

Here, wave factor k=$\frac{\omega }{v}=\frac{6\times {10}^8}{3\times {10}^8}=2m^{-1}$

As E=cB

so required ratio$$\begin{gathered} \frac{\mathrm{E}}{\mathrm{B}}=\mathrm{c} \\ \frac{\mathrm{B}}{\mathrm{E}}=\frac{1}{\mathrm{c}} \end{gathered}$$

Therefore, the ratio of amplitude of magnetic field

to the amplitude of electric field for an electromagnetic

wave propagating in vacuum is equal to reciprocal

of speed of light.