question_answer

The oscillating electric field is

\[{{E}_{y}}=30\sin \,[2\times {{10}^{11}}t+300\pi x]\,V{{m}^{-1}}\]

(i) Obtain the value of the wavelength of the electromagnetic wave.

(ii) Write down the expression for the oscillating magnetic field.

Answer:

Comparing the equation of oscillating electric field with the standard wave equation, \[E(x,t)={{E}_{0}}\sin (\omega t+kx)\] We find that propagation constant \[k=300\pi \] and amplitude of oscillating electric field \[{{E}_{0}}=30\,V{{m}^{-1}}\] \[\therefore \] Wavelength of the electromagnetic wave,             \[\lambda =\frac{2\pi }{k}=\frac{2\pi }{300\pi }=\frac{1}{150}m\] (ii) The electromagnetic wave is propagating along negative direction of X-axis and electric field is oscillating along Y-axis. As the direction of wave propagation is the direction of \[\mathbf{E\times B,}\] hence magnetic field must be oscillating along negative Z-axis, moreover, its amplitude \[{{B}_{0}}=\frac{{{E}_{0}}}{c}=\frac{30}{3\times {{10}^{8}}}={{10}^{-7}}\,T\] Hence, expression for oscillating magnetic field is \[{{B}_{z}}={{10}^{-7}}\sin \] \[[2\times {{10}^{11}}t+300\pi x]\,T\]