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NEET-XII-Physics

40: Electromagnetic Waves

with Solutions - page 3

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  • Qstn #5
    Using B = µ0 H, find the ratio E0/H0 for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.
    Ans : Given, B = µ0H
    For vacuum we can rewrite this equation as,
    B0 = µ0H0 ...(i)
    Relation between magnetic field and electric field for vacuum is given as,
    B0 = µ0`` {\in }_{0}``cE0 ...(ii)
    From equation (i) by (ii),
    `` {\,\mathrm{\,\mu \,}}_{0}{H}_{0}={\,\mathrm{\,\mu \,}}_{0}{\in }_{0}c{E}_{0}``
    `` \Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{{\in }_{0}c}``
    `` \Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{8.85\times {10}^{-12}\times 3\times {10}^{8}}``
    `` \Rightarrow \frac{{E}_{0}}{{H}_{0}}\approx 377\,\mathrm{\,\Omega \,}``
    `` \,\mathrm{\,Dimension\,}\text{of}\frac{1}{{\in }_{0}c}=\frac{1}{\left[{\,\mathrm{\,LT\,}}^{-1}\right]\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]}``
    `` =\frac{1}{{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-2}{\,\mathrm{\,T\,}}^{3}{\,\mathrm{\,A\,}}^{2}}``
    `` ={\,\mathrm{\,M\,}}^{4}{\,\mathrm{\,L\,}}^{2}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-2}=\left[R\right]``
    Page No 339:
  • Qstn #6
    The sunlight reaching Earth has maximum electric field of 810 Vm-1. What is the maximum magnetic field in this light?
    Ans : Given:
    Electric field amplitude, E0 = 810 V/m
    Maximum value of magnetic field = Magnetic field amplitude = B0 = ?
    We know:
    Speed of a wave =`` \frac{E}{B}``
    For electromagnetic waves, speed = speed of light
    B0 = µ0 ε0 cE0
    Putting the values in the above relation, we get:
    `` {B}_{0}=4\,\mathrm{\,\pi \,}\times {10}^{-7}\times 8.85\times {10}^{-12}\times 3\times {10}^{8}\times 810``
    `` {B}_{0}=4\times 3.14\times 8.85\times 3\times 81\times {10}^{-10}``
    `` {B}_{0}=27010.9\times {10}^{-10}``
    `` {B}_{0}=2.7\times {10}^{-6}\,\mathrm{\,T\,}=2.7\,\mathrm{\,\mu T\,}``
    Page No 339:
  • Qstn #7
    The magnetic field in a plane electromagnetic wave is given by
    B = (200 µT) sin [(4.0 × 1015s-1)(t-x/c)].
    Find the maximum electric field and the average energy density corresponding to the electric field.
    Ans : Maximum value of a magnetic field, B0 = 200 `` \,\mathrm{\,\mu \,}``T
    The speed of an electromagnetic wave is c.
    So, maximum value of electric field,
    `` {E}_{0}=c{B}_{0}``
    `` {E}_{0}=\,\mathrm{\,c\,}\times {B}_{0}=200\times {10}^{-6}\times 3\times {10}^{8}``
    `` {E}_{0}=6\times {10}^{4}{\,\mathrm{\,NC\,}}^{-1}``
    (b) Average energy density of a magnetic field,
    `` {U}_{av}=\frac{1}{2{\,\mathrm{\,\mu \,}}_{0}}{{B}_{0}}^{2}=\frac{(200\times {10}^{-6}{)}^{2}}{2\times 4\,\mathrm{\,\pi \,}\times {10}^{-7}}``
    `` {U}_{av}=\frac{4\times {10}^{-8}}{8\,\mathrm{\,\pi \,}\times {10}^{-7}}=\frac{1}{20\,\mathrm{\,\pi \,}}``
    `` {U}_{av}=0.0159\approx 0.016\,\mathrm{\,J\,}/{\,\mathrm{\,m\,}}^{3}``
    For an electromagnetic wave, energy is shared equally between the electric and magnetic fields.
    Hence, energy density of the electric field will be equal to the energy density of the magnetic field.
    Page No 339:
  • Qstn #8
    A laser beam has intensity 2.5 × 1014 W m-2. Find amplitudes of electric and magnetic fields in the beam.
    Ans : Given:
    Intensity, I = 2.5 × 1014 W/m2
    We know:
    `` I=\frac{1}{2}{\in }_{0}{{E}_{0}}^{2}c``
    `` \Rightarrow {{E}_{0}}^{2}=\frac{2I}{{\mathit{\in }}_{0}c}``
    `` \Rightarrow {E}_{0}=\sqrt{\frac{2I}{{\mathit{\in }}_{0}c}}``
    `` \Rightarrow {E}_{0}=\sqrt{\frac{2\times 2.5\times {10}^{14}}{8.85\times {10}^{-12}\times 3\times {10}^{8}}}``
    `` \Rightarrow {E}_{0}=0.4339\times {10}^{9}``
    `` \Rightarrow {E}_{0}=4.33\times {10}^{8}\,\mathrm{\,N\,}/\,\mathrm{\,C\,}``
    Maximum value of magnetic field,
    `` {B}_{0}=\frac{{E}_{0}}{c}``
    `` {B}_{0}=\frac{4.33\times {10}^{8}}{3\times {10}^{8}}``
    `` \Rightarrow {B}_{0}=1.43\,\mathrm{\,T\,}``
    Page No 339:
  • Qstn #9
    The intensity of the sunlight reaching Earth is 1380 W m-2. Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.
    Ans : Given:
    `` I=1380\,\mathrm{\,W\,}/{\,\mathrm{\,m\,}}^{2}``
    `` I=\frac{1}{2}{\in }_{0}{{E}_{0}}^{2}c``
    `` \Rightarrow {E}_{0}=\sqrt{\frac{2I}{{\in }_{0}c}}``
    `` \Rightarrow {E}_{0}=\sqrt{\frac{2\times 1380}{8.83\times {10}^{-12}\times 3\times {10}^{8}}}``
    `` \Rightarrow {E}_{0}=103.95\times {10}^{4}``
    `` \Rightarrow {E}_{0}=10.195\times {10}^{2}``
    `` \Rightarrow {E}_{0}=1.02\times {10}^{3}\,\mathrm{\,N\,}/\,\mathrm{\,C\,}``
    `` ``
    `` \,\mathrm{\,Since\,}{E}_{0}={B}_{\mathit{0}}c,``
    `` {B}_{0}=\frac{{E}_{0}}{c}=\frac{1.02\times {10}^{2}}{3\times {10}^{28}}``
    `` \Rightarrow {B}_{0}=3.398\times {10}^{-6}``
    `` \Rightarrow {B}_{0}=3.4\times {10}^{-6}\,\mathrm{\,T\,}``
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