Physics M.C.Q (1 Marks)

$\oint B \cdot d l=\mu_0\left(I_C+I_D\right)$

$\text { For Loop } L_1 \quad I_C \neq 0 \text { and } I_D=0$

$\text { For Loop } L_2 \quad I_C=0 \text { and } I_D \neq 0$

$\text { Due to KCL } \quad I_C=I_D$

$i _{ C }= i _{ D }=\left( V _{ O } \omega C \right) \sin \omega t$

On decreasing frequency $i_C \downarrow$

$B _0=\frac{ E _0}{ C }$

$=\frac{48}{3 \times 10^8}$

$=1.6 \times 10^{-7}\,T$

$E=3 \times 10^{-8} \times 3 \times 10^8=9\,N / C$

$\therefore E=9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)$

$(b)$ Microwave $\approx$ (iii) $10^{-2} m$ (iii)

$(c)$ Infrared radiations $\approx$ (iv) $10^{-4} m$ (iv)

$(d)$ $X-\operatorname{ray}( i ) \approx \mathring A \mathring A=10^{-10} m$ (i)

$(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)$

$C=\vec{E} \times \vec{B}$

$(-\hat{j}+\hat{k}) \times(-\hat{j}-\hat{k})$

$=\hat{i}+\hat{i}$

$=2 \hat{i}$

$\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{CdV}}{\mathrm{dt}}$

$\mathrm{I}_{\mathrm{d}}=\mathrm{C}\left(\mathrm{V}_{0} \omega \mathrm{cos} \omega \mathrm{t}\right)$

$=\mathrm{V}_{0} \omega \mathrm{C} \cos \omega \mathrm{t}$

$E = IAt$

$=\frac{20}{10^{-4}} \times 20 \times 10^{-4} \times 60$

$=24 \times 10^{3}\, J$

General equation of magnetic field vector

$B = B _{0} \sin ( k x+\omega t ) T$

$k=\pi \times 10^{3}$

$\frac{2 \pi}{\lambda}=\pi \times 10^{3}$

$\lambda=2 \times 10^{-3} m$

$\mathrm{Q}=\mathrm{CV}$

$\therefore \mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=20 \mu \mathrm{F} \times 3 \mathrm{V} / \mathrm{s}=60\; \mu \mathrm{A}$

Also, conduction current in wires is equal to displacement current between the plates of capacitor.

$\therefore \quad v \hat{i}=(E \hat{j}) \times(\vec{B})$

$[\because \vec{E}=E \hat{j}(\text { Given })]$

As $\hat{i}=\hat{j} \times \hat{k}$ so $\vec{B}=B \hat{k}$

Direction of oscillating magnetic field of the em wave will be along $+z$ direction.

$\frac{E_{\mathrm{rms}}}{B_{\mathrm{rms}}}=c$ or $B_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{c}$

$B_{\mathrm{rms}}=\frac{6}{3 \times 10^{8}}=2 \times 10^{-8}\, \mathrm{T}$

since, $B_{\mathrm{rms}}=\frac{B_{0}}{\sqrt{2}}$

where $B_{0}$ is the peak value of magnetic field

$\therefore B_{0} =B_{\mathrm{rms}} \sqrt{2}=2 \times 10^{-8} \times \sqrt{2} T $

$ B_{0}  \approx 2.83 \times 10^{-8}\, \mathrm{T}$

Net impedance, $Z=\sqrt{R^{2}+X_{L}^{2}}=100 \sqrt{2}\, \Omega$

Peak value of displacement current

$=$ Maximum conduction current in the circuit

$=\frac{\varepsilon_{0}}{Z}=\frac{220 \sqrt{2}}{100 \sqrt{2}}=2.2 \,\mathrm{A}$

$=\mathrm{C} \frac{\mathrm{dv}}{\mathrm{dt}}$

$=2 \times 10^{-6} \times 10^{6}$

$=2 \mathrm{\,A}$

$I_d=\varepsilon_0 A \frac{d E}{d t}=\varepsilon_0\left(\pi R^2\right) \frac{d E}{d t}=i$

$I_d^{\prime}=\varepsilon_0\left(\pi R^2-\frac{\pi R^2}{4}\right) \frac{d E}{d t}$

$=\frac{3}{4} \varepsilon_0 \pi R^2 \frac{d E}{d t}$

$I_d=\frac{3}{4} i$

$I_d=\frac{A \varepsilon_0}{d} \frac{d E}{d t} \times d$

$I=\frac{1}{2} \frac{d E}{d t} \times d$

$I=\frac{1}{2} \frac{d V}{d t} \times \frac{1}{d} \cdot d$

$\frac{d V}{d t}=2 I$

$=\frac{h}{\lambda}=\frac{h}{c / v}=\frac{h v}{c}=\frac{E}{c}$

Momentum over unit area

$=\frac{E}{A c}=\frac{I}{c}\left[I=\frac{E}{A} \text { for wave }\right]$

since surface is non reflecting, final momentum of photon $= 0,$ change in momentum $=\frac{I}{c}$

So, force per unit area $=\frac{I}{c}$

Pressure of radiation $=\frac{I}{c}$

$=$$1.6 \times {10^{ - 10}}kg - m/{s^2}$

${{\text{i}}_{\text{d}}} =  \in { \in _0}\frac{{{\text{d}}{\phi _{\text{E}}}}}{{{\text{dt}}}} =  \in { \in _0}{\text{A}}\frac{{{\text{dE}}}}{{{\text{dt}}}}$

$ =  \in { \in _0} \times {\text{A}}{{\text{E}}_0}\omega \cos (\omega {\text{t}} - kx)$

${\rm{Jd}} =  \in { \in _0}{{\rm{E}}_0}\omega $

$\frac{{{{\rm{J}}_{\rm{e}}}}}{{{{\rm{J}}_{\rm{d}}}}} = \frac{\sigma }{{ \in { \in _0}\omega }}$

$\mathrm{q} \mathrm{E} \cdot \mathrm{t}=\mathrm{mv}_{0} \sin \theta$

$E_{y}=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$

Therefore, for the magnetic field in $Z$ - direction

$\mathrm{B}_{\mathrm{z}}=\frac{\mathrm{E}_{\mathrm{y}}}{\mathrm{c}}$

$=\left(\frac{66}{3 \times 10^{8}}\right) \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)$

$=22 \times 10^{-8} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$

$=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)$    

$\mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\in_{0} \mathrm{c}}}$ or $\sqrt{\frac{2 \times 500 \times 10^{9} \times 36 \pi}{\pi \times 3 \times 10^{8}}}$

$\mathrm{E}_{0}=2 \sqrt{3} \times 10^{2} \mathrm{\,N} / \mathrm{C}$

${E_{0}=\sqrt{\frac{P}{2 \pi R^{2} \varepsilon_{0} c}}} $

$ {=\sqrt{\frac{800}{2 \times 3.14 \times(4)^{2} \times 8.86 \times 10^{-12} \times 3 \times 10^{8}}}}$

$ {=54.77 \mathrm{\,V} / \mathrm{m}}$

$\overrightarrow{\mathrm{F}}=-e(\overrightarrow{\mathrm{E}}+\vec{v} \times \overrightarrow{\mathrm{B}})$

The electron will be undeflected if $\mathrm{v} \perp \mathrm{B}.$

If $\mathrm{E}$ is along $+\mathrm{z}$ -direction, the force $-\mathrm{e}$ $\mathrm{E}$ will be along $\mathrm{z}$ -direction. If $\mathrm{B}$ is along $+\mathrm{x}$ direction, force

$-\mathrm{e}(\mathrm{v} \times \mathrm{B})$ will be along $+\mathrm{z}$ divection. When $\mathrm{eE}$ $=$ $\mathrm{evB}.$ the electron moves along $+\mathrm{y}$ -direction undeflected. Hence the correct choice is $(3).$ 

Thus, for an electron moving along $+$ $\mathrm{y}$ direction. the electric field should be along $+\mathrm{z}$ direction and magnetic field along $+\mathrm{x}$ direction, then the electron will be undeflected.