Electromagnetic Waves Test

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Electromagnetic Waves Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is the right derivation for Ampere-Maxwell law?

A
$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + 4\pi {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{dt}$$

B
$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + 3 {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{idt}$$

C
$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{dt}$$

D
$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + \dfrac{{\mu_0}}{ {\varepsilon_0}} \dfrac{d {\phi}_E}{dt}$$

Solution

Maxwell pointed out the inconsistency of the Ampere's circuital law. He said that there is also a current flow, in the space between the plates of a capacitor due to time varying electric field between the plates. Maxwell named this current as displacement current ($$i_{d}$$). This concept makes a continuous flow of current in an electric circuit containing a capacitor.
Now the improved Ampere's law, called Ampere-Maxwell's law is given by:
$$\oint\vec B.\vec dl=\mu_{0}i_{c}+\mu_{0}\varepsilon_{0}\dfrac{d\phi_{E}}{dt}$$
Question 2
1 / -0

What is the frequency of electromagnetic waves in a vacuum that have the same wavelength as a 500.0 Hz sound wave moving at 345 m/s?

A
500.0 Hz

B
0.690 Hz

C
$$2.30 \times 10^{9} \, Hz$$

D
1.45 Hz

E
$$4.35 \times 10^8 \, Hz$$

Solution

For sound waves : $$\nu = 500.0$$ Hz $$v = 345$$ m/s
Wavelength of sound waves $$\lambda_s =\dfrac{v}{\nu} = \dfrac{345}{500.0} = 0.69$$ m

According to the question, wavelength of EM waves $$\lambda_E = \lambda_s = 0.69$$ m
Speed of EM waves $$v' = 3\times 10^8$$ m/s
Thus frequency of EM waves $$\nu'= \dfrac{v'}{\lambda_E} = \dfrac{3\times 10^8}{0.69} =4.35 \times 10^8 $$ Hz
Question 3
1 / -0

Which of the following statement(s) is/are correct?

A
Conduction current obeys Ohm's law whereas displacement current does not.

B
Conduction current is the actual current whereas displacement current is the apparent

current produced by time varying electric field.

C
Conduction current density is represented by $$\overrightarrow {J_c} = \sigma \overrightarrow E$$ whereas displacement current

density is given by $$ \overrightarrow {J_d} = \dfrac{ \overrightarrow{\delta E}}{\delta t}$$; $$\sigma $$= conductivity of the element, $$\overrightarrow E$$= electric field.

D
All of the above

Solution

Displacement current is the current that occurs due to charging electric field introduced by maxwell. It depends on the frequency of electric field while conduction current follows ohms law, requires medium displacement.Current doesnot follow ohms law nor require medium.
So option "Conduction current obeys Ohm's law whereas displacement current does not" is correct
Question 4
1 / -0

What is the frequency of electromagnetic waves in a vacuum that have the same wavelength as a $$500.0\ Hz$$ sound wave moving at $$345\ m/s$$?

A
$$500.0\ Hz$$

B
$$0.690\ Hz$$

C
$$2.30\times 10^{-9}\ Hz$$

D
$$1.45\ Hz$$

E
$$4.35\times 10^{8}\ Hz$$

Solution

We know the relation $$v=\lambda\nu$$
Hence for waves with same wavelength,
$$\dfrac{v_1}{\nu_1}=\dfrac{v_2}{\nu_2}$$
$$\implies \nu_2=\dfrac{v_2}{v_1}\nu_1$$
$$=\dfrac{3\times 10^8}{345}\times 500Hz=4.35\times 10^8Hz$$
Question 5
1 / -0

Radiations of intensity 0.5 W/m2 are striking a metal plate. The pressure on the plate is

A
$$0.166 \times 10^{-8} N/m2$$

B
$$0.332 \times 10-8 N/m^2$$

C
$$0.111 \times 10-8 N/m^2$$

D
$$0.083 \times 10-8 N/m^2$$

Solution

Intensity or power per unit area of the radiations,
$$P = pv$$
$$\Rightarrow p=\dfrac{p}{v} = \dfrac{0.5}{3\times 10^8}=0.166\times 10^{-8}N/m^2$$
Question 6
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A $$1000\Omega $$ resistance and a capacitor of $$100\Omega $$ resistance are connected in series a $$220 V$$ source. when the capacitor is 50% charged, the value of the displacement current is.

A
$$11.\sqrt{2}A$$

B
$$2.2 A$$

C
$$11A$$

D
$$4.4A$$

Solution

Displacement current $$= I_D=C\dfrac{dV}{dT}=C\omega V_o=\dfrac{V_o}{X_c}=\dfrac{220V}{100\Omega}=2.2A$$
As we are asked amplitude of displacement current. So, we don't have to worry about charge on capacitor.
Question 7
1 / -0

A parallel plate capacitor having plate area A and plate separation $$d$$ is connected to a battery of emf $$\varepsilon$$ and internal resistance $$R$$ at $$t=0$$. Consider a plane surface of area $$\dfrac{A}{2}$$, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time?

A
$$\dfrac {\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$$

B
$$\dfrac {2\varepsilon}{R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$$

C
$$\dfrac {5\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4 \varepsilon AR}}$$

D
$$\dfrac {\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4\pi \varepsilon AR}}$$

Solution
Question 8
1 / -0

A parallel plate capacitor consists of circular plates with radius $$10 cm$$ , separated by a distance of $$0.5mm$$. The capacitor is charged by an external source such that the electric field between the plates changes at a rate $$5 \times 10^{13} Vm{-1}s{-1}$$. The displacement current through the capacitor is

A
$$1.4 mA$$

B
$$14 mA$$

C
$$14 A$$

D
$$1.4 A$$

Solution

Displacement current $$=\dfrac{dQ}{dt}=\dfrac{d(CV)}{dt}$$
$$=C\dfrac{dV}{dt}\\ =C(d.\dfrac{dE}{dt})$$
$$=\dfrac{A\epsilon_0}{d}.d\dfrac{dE}{dt}$$
$$=A\epsilon_0\dfrac{dE}{dt}$$
$$=\pi r^2 \epsilon_0\dfrac{dE}{dt}$$
$$=14A$$
Question 9
1 / -0

A plane electromagnetic wave of frequency 20 MHz travels through a space along x direction. If the electric field vector at a certain point in space is 6 V $$m^{-1}$$, what is the magnetic field vector at that point?

A
$$2\times 10^{-8}T$$

B
$$\dfrac{1}{2}\times 10^{-8}T$$

C
$$2T$$

D
$$\dfrac{1}{2}T$$

Solution

Velocity of EM wave $$v = 3\times 10^{8}$$ $$m/s$$
Electric field vector $$E = 6$$ $$V/m$$
Thus magnetic field vector $$B = \dfrac{E}{v}$$
$$\therefore$$ $$B = \dfrac{6}{3\times 10^8} =2\times 10^{-8}$$ $$T$$
Question 10
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A parallel plate capacitor with plate area A and separation between the plates $$d$$, is charged by a constant current $$i$$. Consider a plane surface of area $$\dfrac{A}{2}$$ parallel to the plate and down symmetrically between the plates. Find the displacement current through this area.

A
$$i_d= \dfrac{i}{8}$$

B
$$i_d= \dfrac{i}{2}$$

C
$$i_d= \dfrac{i}{3}$$

D
$$i_d= \dfrac{i}{4}$$

Solution

Displacement current $$=\dfrac{dQ}{dt}=\dfrac{d(CV)}{dt}$$
$$=C\dfrac{dV}{dt}=C(d.\dfrac{dE}{dt})$$
$$=\dfrac{A\epsilon_0}{d}.d\dfrac{dE}{dt}$$
$$=A\epsilon_0\dfrac{dE}{dt}\propto A$$
Hence, current halves when area of plates halves.

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Electromagnetic Waves Test - 21

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Electromagnetic Waves Test - 21

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Question 1

$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + 4\pi {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{dt}$$

$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + 3 {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{idt}$$

$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + {\mu_0} {\varepsilon_0} \dfrac{d {\phi}_E}{dt}$$

$$\oint {\overrightarrow B}. \overrightarrow {dl} = {\mu}_0 i_c + \dfrac{{\mu_0}}{ {\varepsilon_0}} \dfrac{d {\phi}_E}{dt}$$

Solution

Question 2

500.0 Hz

0.690 Hz

$$2.30 \times 10^{9} \, Hz$$

1.45 Hz

$$4.35 \times 10^8 \, Hz$$

Solution

Question 3

Conduction current obeys Ohm's law whereas displacement current does not.

Conduction current is the actual current whereas displacement current is the apparentcurrent produced by time varying electric field.

All of the above

Solution

Question 4

$$500.0\ Hz$$

$$0.690\ Hz$$

$$2.30\times 10^{-9}\ Hz$$

$$1.45\ Hz$$

$$4.35\times 10^{8}\ Hz$$

Solution

Question 5

$$0.166 \times 10^{-8} N/m2$$

$$0.332 \times 10-8 N/m^2$$

$$0.111 \times 10-8 N/m^2$$

$$0.083 \times 10-8 N/m^2$$

Solution

Question 6

$$11.\sqrt{2}A$$

$$2.2 A$$

$$11A$$

$$4.4A$$

Solution

Question 7

$$\dfrac {\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$$

$$\dfrac {2\varepsilon}{R} \ \ \ e^{\dfrac{-td}{\varepsilon AR}}$$

$$\dfrac {5\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4 \varepsilon AR}}$$

$$\dfrac {\varepsilon}{2R} \ \ \ e^{\dfrac{-td}{4\pi \varepsilon AR}}$$

Solution

Question 8

$$1.4 mA$$

$$14 mA$$

$$14 A$$

$$1.4 A$$

Solution

Question 9

$$2\times 10^{-8}T$$

$$\dfrac{1}{2}\times 10^{-8}T$$

$$2T$$

$$\dfrac{1}{2}T$$

Solution

Question 10

$$i_d= \dfrac{i}{8}$$

$$i_d= \dfrac{i}{2}$$

$$i_d= \dfrac{i}{3}$$

$$i_d= \dfrac{i}{4}$$

Solution

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Conduction current is the actual current whereas displacement current is the apparent

current produced by time varying electric field.