Ncert Exercises & Solutions

A plane EM wave travelling along z-direction is given by $\vec{E} = E_{0} \sin (kz - ωt) \hat{i} and \vec{B} = B_{0} \sin (kz - ωt) \hat{j} . Show that :$

$(i) Evaluate \oint \vec{E} . d \vec{l} over the rectangular loop 1234 shown in figure .$
$(ii) Evaluate \oint \vec{B} . d \vec{s} over the surface bounded loop 1234 .$
$(iii) Use equation \oint \vec{E} . d \vec{l} = - \frac{{dϕ}_{B}}{dt} to prove \frac{E_{o}}{B_{o}} = c .$
$(iv) By using \oint \vec{B} . d \vec{s} = μ_{o} I + ε_{o} \frac{{dϕ}_{E}}{dt}, prove that c = \frac{1}{\sqrt{μ_{o} ε_{o}}} .$

Hint: The electromagnetic wave propagates perpendicular to both the electric field and the magnetic field.

(I) Step 1: Consider the figure given below.

During the propagation of electromagnetic wave along z-axis, let electric field vector

\vec{E}

be along x-axis and magnetic field vector

\vec{B}

along the y-axis, i.e.,

\vec{E} = E_{0} \hat{i} and \vec{B} = B_{0} \hat{i} .

Line integral of

\vec{E}

over the closed rectangular path 1234 in the x-z plane of the figure;

\oint E \cdot dl = \int_{1}^{2} E \cdot dl + \int_{2}^{3} E \cdot dl + \int_{3}^{4} E \cdot dl + \int_{4}^{1} E \cdot dl

= \int_{1}^{2} Edlcos 90 ° + \int_{2}^{3} Edlcos 0 ° + \int_{3}^{4} Edlcos 90 ° + \int_{4}^{1} Edlcos 180 °

= E_{0} h [\sin ({kz}_{2} - ωt) - \sin ({kz}_{1} - ωt)]

(ii) Step 2 : For evaluating \int B \cdot ds, let us consider the rectangle 1234 to be madeof strips of are ds = hdz each .

\int B \cdot ds = \int B \cdot ds \cos 0 ° = \int B \cdot ds = \int_{z_{1}}^{z_{2}} B_{0} \sin (kz - ωt) hdz

= \frac{- B_{0} h}{k} [\cos ({kz}_{2} - ωt) - \cos ({kz}_{1} - ωt)]

(iii) Step 3 : Given, \oint E \cdot dl = \frac{- {dϕ}_{e}}{dt} = - \frac{d}{dt} \oint B \cdot ds

Putting the values from Eqs . (i) and (ii), we get;

E_{0} h [\sin ({kz}_{2} - ωt) - \sin ({kz}_{1} - ωt)]

= \frac{- d}{dt} [\frac{B_{0} h}{k} {\cos ({kz}_{2} - ωt) - \cos ({kz}_{1} - ωt)]

= \frac{B_{0} h}{k} ω [\sin ({kz}_{2} - ωt) - \sin ({kz}_{1} - ωt)]

\Rightarrow E_{0} = \frac{B_{0} ω}{k} = B_{0} c

\Rightarrow \frac{E_{0}}{B_{0}} = c

(iv) Step 4 : For evaluating \int B \cdot dl, let us consider a loop 1234 in y - z plane as shown in figure given below .

\oint B \cdot dl = \int_{1}^{2} B \cdot dl + \int_{2}^{3} B \cdot dl + \int_{3}^{4} B \cdot dl + \int_{4}^{1} B \cdot dl

= \int_{1}^{2} Bdlcos 0 ° + \int_{2}^{3} Bdlcos 90 ° + \int_{3}^{4} Bdlcos 180 ° + \int_{4}^{1} Bdlcos 90 °

= B_{0} h [\sin ({kz}_{1} - ωt) - \sin ({kz}_{2} - ωt)] . . . (iii)

Now to evaluate ϕ_{E} = \int E . ds, let us consider the rectangle 1234 to be made of strips of area {hd}_{2} each .

ϕ_{E} = \int E \cdot ds = \int Edscos 0 ° = \int Eds = \int_{z_{1}}^{z} E_{0} \sin ({kz}_{1} - ωt) hdz

= - \frac{E_{0} h}{k} [\cos ({kz}_{2} - ωt) - \cos ({kz}_{1} - ωt)]

∴ \frac{{dϕ}_{ε}}{dt} = \frac{E_{0} hω}{k} [\sin ({kz}_{1} - ωt) - \sin ({kz}_{2} - ωt)]

In \oint B \cdot dl = μ_{0} (l + ε_{0} \frac{{dϕ}_{E}}{dt}), I = conduction current = 0 in vacuum

∴ \oint B \cdot dl = μ_{0} ε_{0} \frac{{dϕ}_{E}}{dt}

Using relations obtained in Eqs . (iii) and (iv) and simplifying, we get;

B_{0} = E_{o} \frac{{ωμ}_{0} ε_{0}}{k}

\Rightarrow \frac{E_{0}}{B_{0}} \frac{ω}{K} = \frac{1}{μ_{0} ε_{0}}

But, \frac{E_{0}}{B_{0}} = c and ω = ck

\Rightarrow c . c = \frac{1}{μ_{0} ε_{0}}, therefore c = \frac{1}{\sqrt{μ_{0} ε_{0}}}

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