3 Marks Question

Question 13 Marks

Derive the equation of energy density for electromagnetic wave.

Answer

→Energy density : "Energy stored per unit volume is called Energy density." An electromagnetic wave contains Electric field and magnetic field both.
→Energy density associated with Electric field, $\varrho_{ E }=\frac{1}{2} \varepsilon_0 E ^2$
→Energy-density associated with magnetic field, $\varrho _{ B }=\frac{ B ^2}{2 \mu_0}$
→Total Energy-density associated with EM wave,
$\varrho =\varrho_{ E }+\varrho_{B}
\therefore =\frac{1}{2} \varepsilon_0 E ^2+\frac{ B ^2}{2 \mu_0}$
→But the magnitudes of electric field and magnetic field change as per sine or cosine functions in an EM wave.Hence, in the equation, the rms value of electric field and magnetic field is considered.
→Energy-density $ \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{ B _{\text {rms }}^2}{2 \mu_0}$
But $c ^2=\frac{1}{\mu_0 \varepsilon_0} \therefore \mu_0=\frac{1}{ c ^2 \varepsilon_0}$
and $\frac{E_{r m s}}{B_{r m s}}=c \therefore B_{r m s}=\frac{E_{r m s}}{c}$
→Substituting both these values in eq. (1),
$\begin{array}{l}\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{\frac{ E _{\text {rms }}^2}{ c ^2}}{2\left(\frac{1}{ c ^2 \varepsilon_0}\right)} \\\therefore \varrho=\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2+\frac{1}{2} \varepsilon_0 E _{\text {rms }}^2 \\\therefore \varrho=\varepsilon_0 E _{\text {rms }}^2\end{array}$
→In similar manner, $\varrho=\frac{B_{\text {rms }}^2}{\mu_0}$ can also be derived.

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Question 23 Marks

On the basis of the charging of capacitor, clarify that there is something missing in the Ampere circuital law.

Answer

→As shown in fig. (a), a parallel plate capacitor is connected to a time-varying electric current. Magnetic field at a point P outside the parallel plates is to be found out.
→For this, consider a plane circular loop of radius r, whose plane is perpendicular to the direction of the current carrying wire, and which is centred symmetrically with respect to the wire.
→From symmetry, the magnetic field is directed along the circumference of the circular loop and its magnitude is same at all points on the loop.

→From Ampere's circuital law,
$\oint \overrightarrow{ B } \cdot \overrightarrow{d l}=\mu_0 i(t)$
$\begin{array}{l}\oint B d l \cos 0^{\circ}=\mu_0 i(t) \\ \therefore \quad B \oint d l=\mu_0 i(t) \\ \text { but } \oint d l=2 \pi r =\text { circumference of the closed loop } \\ \therefore B (2 \pi r )=\mu_0 i(t)......(1)\end{array}$
→As shown in fig. (b), now think of another surface. This is a pot like surface, which no where touches the current, but has its bottom between the capacitor plates, its mouth is the circular loop mentioned above.

→As shown in fig. (c), think of a third surface which is shaped like a tiffin box without the lid.
→Electric current isn't passing through the surfaces shown in fig. (b) and (c). Hence Ampere's circuital law is applied to such surfaces having different shapes but same perimeter. It gives:
$B \oint \overrightarrow{d l}=\mu_0 i(t) \quad$ (Ampere is circuital law)
→In this equation LHS (left hand side) has not changed (and is not zero), but the RHS (right hand side) becomes zero, since the current enclosed by these loops is zero.
→This is a contradiction; because, calculated one way, there is a magnetic field at point P; calculated another way, the magnetic field at point P is zero.
→Due to this contradicton, we can say that there is something missing in the Ampere circuital law.

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Question 33 Marks

Derive the equation of the missing term in the Ampere circuital law OR Derive the equation of displacement current and write its unit and definition.

Answer

→As shown in the fig. applying Ampere's circuital law to different surfaces, to find out the magnetic field at a point P outside a parallel plate capacitor, magnetic field is found to be different.

→This contradiction indicates that there is something missing in the Ampere circuital law. Hence the following correction needs to be done in the law:
→Suppose, the plates of the capacitor have an area A, and charge Q on each of them.
→Electric field in the region between two plates of the capacitor,
$\begin{aligned} E =\frac{\sigma}{\varepsilon_0} & =\frac{ Q }{\varepsilon_0 A} \\ & \left(\because \sigma=\frac{ Q }{ A } \text { surface charge density }\right)\end{aligned}$
→This field is perpendicular to the surface S of above fig. (c). It has the same magnitude over the area A of the capacitor plates, and it vanishes in the outside region.
→Electric flux associated with surface S,
$\begin{aligned} \phi_{ E } & =\overrightarrow{ E } \cdot \overrightarrow{ A }= EA \cos \theta \text { but } \overrightarrow{ E } \| \overrightarrow{ A } \therefore \theta=0^{\circ} \\ \phi_{ E } & = EA \quad\left(\because \cos 0^{\circ}=1\right) \\ \therefore \quad \phi_{ E } & =\frac{ Q }{\varepsilon_{ o } A } \cdot A =\frac{ Q }{\varepsilon_{ o }}......(1)\end{aligned}$
→Here the charging condition of capacitor is shown, therefore charge Q on the capacitor plates changes with time, and corresponding current will be $i=\frac{d Q }{d t}$.
→Taking the differentiation of eq. (1) with respect to time,
$\begin{array}{l}\therefore \frac{d \phi_{ E }}{d t}=\frac{1}{\varepsilon_{ o }} \cdot \frac{d Q }{d t}=\frac{i}{\varepsilon_0} \\\therefore i=\varepsilon_0 \frac{d \phi_{ E }}{d t}\end{array}$
→This term is the missing term in Ampere's circuital law. This current is known as displacement current $\left(i_d\right)$. Its unit is : C/s or A.
→Definition of Displacement Current : "In some region, the electric current created / induced due to a time-varying electric field ( $\vec{E})$ and hence time-varying electric flux $\left(\phi_E\right)$ is known as displacement current."

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