12th Class Physics Sample Paper Physics Sample Paper-7

  • 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.


    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 \[{{B}_{z}}={{10}^{-7}}\sin \] \[[2\times {{10}^{11}}t+300\pi x]\,T\]

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