Electromagnetic Waves Test

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Electromagnetic Waves Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The electric field intensity at a point in vacuum is equal to

A
Zero

B
Force a proton would experience there

C
Force an electron would experience there

D
Force a unit positive charge would experience there

Solution

The electric field intensity at a point in a vacuum is equal to force experienced by a unit positive charge placed at that point.
Question 2
1 / -0

Displacement current goes through the gap between the plates of a capaitor when the charge of the capacitor:
(i) does not change
(ii) change with time
(iii) increases
(iv) decreases

A
( i ), ( iii ) and ( iv )

B
( i ), and ( ii ) and ( iv )

C
( ii ), ( iii ) and ( iv )

D
( i ), ( ii ) and ( iii )

Solution

Displacement current inside a capacitor
$$I_D=\epsilon_0\dfrac {d\phi_E}{dt}$$
where $$\phi_E$$ is the electric flux inside the capacitor
Up to the time the electric flux changes, there will be a displacement current. This is possible when the charge on a capacitor changes. Therefore, the displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor or the electric field increases or decreases inside the capacitor.
Question 3
1 / -0

A parallel plate condenser consists of two circular plates each of radius $$2cm$$ separated by a distance of $$0.1mm$$. A time varying potential difference of $$5\times {10}^{13}v/s$$ is applied across the plates of the condenser. The displacement current is:

A
$$5.50A$$

B
$$5.56\times {10}^{2}A$$

C
$$5.56\times {10}^{3}A$$

D
$$2.28\times {10}^{4}A$$

Solution

Given: A parallel plate condenser consists of two circular plates each of radius 2cm separated by a distance of 0.1mm. A time varying potential difference of $$5\times 10^{13}v/s$$ is applied across the plates of the condenser.
To find the displacement current
Solution:
Radius of each circular plate, $$r = 2 cm = 0.02 m$$
Distance between the plates, $$d =0.1mm=0.1\times10^{-3}m$$
the change in potential difference between the plates, $$\dfrac {dV}{dt}=5\times10^{13}vs$$
Permittivity of free space, $$\varepsilon_0= 8.85 \times 10^{−12} C^2 N^{−1} m^{−2}$$
Hence area of the each plate, $$A=\pi r^2=\pi (0.02)^2=1.26\times 10^{-3}m^2......(i)$$
Capacitance between the two plates is given by the relation,
$$C=\dfrac {\varepsilon_0A}{d}\\\implies C=\dfrac {8.85\times10^{-12}\times1.26\times10^{-3}}{0.1\times10^{-3}}\\\implies C=11.15\times10^{-11}F$$
Now we know
displacement current,
$$I_d=\dfrac {dq}{dt}$$
where $$q$$ is the charge on each plate and it is equal to $$q=CV$$
$$I_d=C\dfrac {dV}{dt}$$
By substituting the values, we get
$$I_d=11.15\times10^{-11}\times5\times10^{13}\\\implies I_d=57.55\times10^2\\\implies I_d=5.7\times10^3A$$
is the displacement current.
Question 4
1 / -0

Electromagnetic wave of intensity $$1400\ W/m^{2}$$ falls on metal surface on area $$1.5\ m^{2}$$ is completely absorbed by it. Find out force exerted by beam.

A
$$14\times 10^{-5}N$$

B
$$14\times 10^{-6}N$$

C
$$7\times 10^{-5}N$$

D
$$7\times 10^{-6}N$$

Solution

For a perfectly absorbing surface,
$$F=\dfrac{IA}{C}$$

$$=\dfrac{(1400w/m^2\times 1.5m^2)}{(3\times 10^8m/s)}$$

$$=7\times 10^{-6}N$$.
Question 5
1 / -0

A plane electromagnetic wave of frequency $$28MHz$$ travels in free space along the positive x-direction. At a particular point in space and time, electric field is $$9.3V/m$$ along positive y-direction. The magnetic field (in T) at that point is

A
$$3.1\times10^{-8} $$ along positive z-direction

B
$$3.1\times10^{-8} $$ along negative z-direction

C
$$3.2\times10^{7} $$ along positive z-direction

D
$$3.2\times10^{7} $$ along negative z-direction

Solution

$$B = \dfrac{E}{C} = \dfrac{{9.3}}{{3 \times {{10}^8}}} = 3.1 \times {10^-8}$$
Question 6
1 / -0

A parallel plate condenser of capacity $$1 nF$$ is connected to a battery of emf $$2v$$ though a resistance of $$1M \Omega$$. The displacement current after $$10^{-3}\ \sec$$ is:

A
$$0.37\ \mu\ A$$

B
$$0.74\ \mu\ A$$

C
$$1.48\ \mu\ A$$

D
$$0.175\ \mu\ A$$

Solution

$$I=\dfrac{dq}{dt}=\dfrac{v}{R}e^{-t/RC}$$
$$=\dfrac{2}{1\times 10^{6}}e$$
$$=\dfrac{2}{10^6}e^{-1}=\dfrac{2}{10^6}\times 0.3678$$
$$=0.7357 \times 10^{-3}\times 10^{-6}$$
$$=0.7357 \pi A$$
Question 7
1 / -0

At a particular instant, the current in the circuit given below is i. The displacement current between the plates of the capacitor shown below is

A
Zero

B
i

C
$$\frac{i}{2}$$

D
$$\frac{i}{4}$$

Solution
Question 8
1 / -0

A parallel plate condenser has circular plates$$,$$ each of radius $$5cm.$$ It is being charged so that electric field in the gap between its plates rises steadily at the rate of $${10^{12}}V/ms.$$ What is the displacement current$$?$$

A
$$0.07 A$$

B
$$1.39 A$$

C
$$13.9 A$$

D
$$139 A$$

Solution

$$Id= = { \in _0}\dfrac{{d{\phi _E}}}{{dt}}$$

$$ = { \in _0} = \dfrac{{d\left( {EA} \right)}}{{dt}}$$

$$ = { \in _0} = A\dfrac{{dE}}{{dt}}$$

$$ = { \in _0} = \dfrac{{\pi {r^2}}}{{dt}}$$

$$ = 8.85 \times {10^{ - 12}} \times 3.14 \times 25 \times {0^{ - 4}} \times {10^{12}}$$
$$=0.07A$$
Hence,
option $$(A)$$ is correct answer.
Question 9
1 / -0

In an electromagnetic wave,
$$E = 1.2\sin (2\times 10^{-6} t - kx)N/C$$.
Find value of intensity of magnetic field.
$$4\times 10^{-8} field.$$

A
$$4\times {10}^{-8}A/m$$

B
$$0.1\times {10}^{-2}A/m$$

C
$$0.4\times {10}^{-8}A/m$$

D
$$\dfrac {10^{-2}}{\pi}A/m$$

Solution

Given: In an electromagnetic wave, $$E=1.2\sin(2\times10^6t-kx)N/C$$
To find the value of intensity of magnetic field
Solution:
we know the equation of the electromagnetic wave is
$$E=E_0\sin(kx-\omega t)$$
Comparing this with the given equation,
$$E_0=1.2$$
The value of intensity of magnetic field
$$B_0=\dfrac {E_0}{c}\\\implies B_0=\dfrac {1.2}{3\times10^8}\\\implies B_0=0.4\times10^{-8}A/m$$
Question 10
1 / -0

The electric and the magnetic field, associated with an E.M. wave , propagating along the $$+z$$ axis, can be represented by

A
$$[\overrightarrow E=E_o\hat j,\overrightarrow B=B_o\hat k]$$

B
$$[\overrightarrow E=E_o\hat i,\overrightarrow B=B_o\hat j]$$

C
$$[\overrightarrow E=E_o\hat k,\overrightarrow B=B_o\hat i]$$

D
$$[\overrightarrow E=E_o\hat j,\overrightarrow B=B_o\hat i]$$

Solution

Dirft velocity of propagation $$=z$$
So, $$E$$ & $$B$$ should be $$i$$ & $$j$$ ($$x$$ & $$y$$)
as these are perpendicular to each other
$$\overrightarrow E\times \overrightarrow B=+z$$
$$\overrightarrow E$$ must be $$\hat i$$
$$\overrightarrow B$$ must be $$\hat j$$
As, $$\hat i \times \hat j=\hat k$$
and $$\hat j\times \hat i=-\hat k$$

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Electromagnetic Waves Test - 28

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Electromagnetic Waves Test - 28

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Question 1

Zero

Force a proton would experience there

Force an electron would experience there

Force a unit positive charge would experience there

Solution

Question 2

( i ), ( iii ) and ( iv )

( i ), and ( ii ) and ( iv )

( ii ), ( iii ) and ( iv )

( i ), ( ii ) and ( iii )

Solution

Question 3

$$5.50A$$

$$5.56\times {10}^{2}A$$

$$5.56\times {10}^{3}A$$

$$2.28\times {10}^{4}A$$

Solution

Question 4

$$14\times 10^{-5}N$$

$$14\times 10^{-6}N$$

$$7\times 10^{-5}N$$

$$7\times 10^{-6}N$$

Solution

Question 5

$$3.1\times10^{-8} $$ along positive z-direction

$$3.1\times10^{-8} $$ along negative z-direction

$$3.2\times10^{7} $$ along positive z-direction

$$3.2\times10^{7} $$ along negative z-direction

Solution

Question 6

$$0.37\ \mu\ A$$

$$0.74\ \mu\ A$$

$$1.48\ \mu\ A$$

$$0.175\ \mu\ A$$

Solution

Question 7

Zero

i

$$\frac{i}{2}$$

$$\frac{i}{4}$$

Solution

Question 8

$$0.07 A$$

$$1.39 A$$

$$13.9 A$$

$$139 A$$

Solution

Question 9

$$4\times {10}^{-8}A/m$$

$$0.1\times {10}^{-2}A/m$$

$$0.4\times {10}^{-8}A/m$$

$$\dfrac {10^{-2}}{\pi}A/m$$

Solution

Question 10

$$[\overrightarrow E=E_o\hat j,\overrightarrow B=B_o\hat k]$$

$$[\overrightarrow E=E_o\hat i,\overrightarrow B=B_o\hat j]$$

$$[\overrightarrow E=E_o\hat k,\overrightarrow B=B_o\hat i]$$

$$[\overrightarrow E=E_o\hat j,\overrightarrow B=B_o\hat i]$$

Solution

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