M.C.Q (1 Marks)

Question 11 Mark

Dimensions of $\frac{1}{(\mu_0\epsilon_0)}$ is:

$\frac{\text{L}}{\text{T}}$
$\frac{\text{T}}{\text{L}}$
$\frac{\text{L}^2}{\text{T}^2}$
$\frac{\text{T}^2}{\text{L}^2}$

Answer

$\frac{\text{L}^2}{\text{T}^2}$

Explanation:
The speed of light, $\text{C}=\frac{1}{\sqrt{\mu_0\epsilon_0}}$
The dimensions of $\frac{1}{\sqrt{\mu_0\epsilon_0}}$ are of velocity, i.e., $\frac{\text{L}}{\text{T}}$
Therefore, $\frac{1}{\epsilon_0\mu_0}$ will have dimensions $\frac{\text{L}^2}{\text{T}^2}$

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Question 21 Mark

An electromagnetic wave going through vacuum is described by $\text{E}=\text{E}_0\sin(\text{kx}-\omega\text{t})$ which of the following is/are independent of the wavelength?

$\text{k}$
$\omega$
$\frac{\text{k}}{\omega}$
$\text{k}\omega$

Answer

$\frac{\text{k}}{\omega}$

Explanation:
The given quantities can be expressed as,
k is given by,
$\omega$ is given by,
$\omega=2\pi\text{v}$
$\text{c}=\text{v}\lambda$
$\omega=2\pi\frac{\text{c}}{\lambda}$
$\frac{\text{k}}{\omega}$ is given by,
$\frac{\text{k}}{\omega}=\frac{\frac{2\pi}{\lambda}}{\frac{2\pi\text{c}}{\lambda}}=\frac{1}{\text{c}}$
$\text{k}\omega$ is given by,
$\text{k}\times\omega=\frac{2\pi}{\lambda}\times\frac{2\pi\text{c}}{\lambda}=\frac{4\pi^2\text{c}}{\lambda^2}$
Thus, $\frac{\text{k}}{\omega}$ is independent of the wavelength.

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MCQ 31 Mark

An electromagnetic wave going through vacuum is described by $\text{E}=\text{E}_0\sin(\text{kx}-\omega\text{t}),\text{ B}=\text{B}_0\sin(\text{kx}-\omega\text{t})$ Then$:$

✓
$\text{E}_0\text{k}=\text{B}_0\omega$
B
$\text{E}_0\text{B}_0=\omega\text{k}$
C
$\text{E}_0\omega=\text{B}_0\text{k}$
D
None of these.

Answer

Correct option: A.

$\text{E}_0\text{k}=\text{B}_0\omega$

The relation between $E_0$ and $B_0$ id given by $\frac{\text{E}_0}{\text{B}_0}=\text{c}\ ....(\text{i})$
Here, $c =$ Speed of the electromagnetic wave,
The relation between $\omega ($the angular frequency$)$ and $k($wave number$),$
$\frac{\omega}{\text{k}}=\text{c}\ ...(\text{ii})$
Therefore, from $(i)$ and $(ii),$ we get
$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$
$\text{E}_0\text{k}=\text{B}_0\omega$

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Question 41 Mark

A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle:

Does not deflect.
Deflects for a very short time and then comes back to the original position.
Deflects and remains deflected as long as the battery is connected.
Deflects and gradually comes to the original position in a time that is large compared to the time constant.

Answer

Deflects and gradually comes to the original position in a time that is large compared to the time constant.

Explanation:
The compass needle deflects due to the presence of the magnetic field. Inside the capacitor, a magnetic field is produced when there is a changing electric field inside it. As the capacitor is connected across the battery, the charge on its plates at a certain time tis given by,
$\text{Q}=\text{CV}\Big(1-\text{e}^{-\tau/\text{RC}}\Big),$
Q = Charge developed on the plates of the capacitor.
R = Resistance of the resistor connected in series with the capacitor.
C = Capacitance of the capacitor.
V = Potential difference of the battery.
The time constant of the capacitor is given, $\tau=\text{RC}$
The capacitor keeps on charging up to the time $\tau$. The development of charge on the plates will be gradual after $\tau=\text{RC}$ The change in electric field will be up to the time the charge is developing on the plates of the capacitor. Thus, the compass needle deflects and gradually comes to the original position in a time that is large compared to the time constant.

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Question 51 Mark

A magnetic field can be produced by:

A moving charge.
A changing electric field.
Neither of them.
Both of them.

Answer

Both of them.

Explanation:
According to Ampere-Maxwell's Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by,
$\text{i}_\text{d}=\epsilon_0\frac{\text{d}\phi_\text{E}}{\text{dt}}$ $\big(\because\phi_\text{E}$ is the electric flux$\big)$
Thus, the magnetic field is produced by the moving charge as well as the electric field.

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Question 61 Mark

Speed of electromagnetic waves is the same:

For all wavelengths.
In all media.
For all intensities.
For all frequencies.

Answer

For all intensities.

Explanation:
For any given medium, the speed (c) of an electromagnetic wave is given by,
$\text{C}=\text{v}\lambda$
Where,
V = Frequency of the electromagnetic wave.
$\lambda=$ wavelength of the electromagnetic wave.
As the frequency and wavelength are changed, the speed of the electromagnetic wave changes. So, the speed of an electromagnetic wave is not same for all wavelengths and all frequencies in any medium. The velocity of an electromagnetic wave changes with change in medium. Also, the speed of an electromagnetic wave is same for all the intensities in any medium.

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Question 71 Mark

Which of the following have zero average value in a plane electromagnetic wave?

Electric field.
Magnetic field.
Electric energy.
Magnetic energy.

Answer

Electric field.
Magnetic field.

Explanation:
In a plane electromagnetic wave, the electric and the magnetic fields oscillate sinusoidally. For an electromagnetic wave propagating in the z-direction, the electric and magnetic fields are given by,
$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$
$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$
These are sinusoidal functions. Therefore, for a fixed value of z. the average value of the electric and magnetic fields are zero.

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MCQ 81 Mark

Consider the following two statements regarding a linearly polarised, plane electromagnetic wave$:$

The electric field and the magnetic field have equal average values.
The electric energy and the magnetic energy have equal average values.

A
Both $A$ and $B$ are true.
B
$A$ is false but $B$ is true.
✓
$B$ is false but $A$ is true
D
Both $A$ and $B$ are false.

Answer

Correct option: C.

$B$ is false but $A$ is true

For a linearly polarised, plane electromagnetic wave,
$\text{E}=\text{E}_0\sin\omega\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$\text{B}=\text{B}_0\sin\omega\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
The average value of either $E$ or Bover a cycle is zero $($average of $\sin(\theta)$ over a cycle is zero$).$
Also the electric energy density $(U_E)$ and magnetic energy density $(U_B)$ are equal.
$\text{u}_\text{E}=\frac{1}{2}\in_0\text{E}^2=\frac{\text{B}^2}{2\mu_0}=\text{u}_\text{B}$
Energy can be found out by integrating energy density over the entire volume of full space.
As the energy of the electromagnetic wave is equally shared between electric and magnetic field so their average values will also be equal.

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Question 91 Mark

An electric field $\overrightarrow{\text{E}}$ and a magnetic field $\overrightarrow{\text{B}}$ exist in a region. The fields are not perpendicular to each other.

This is not possible.
No electromagnetic wave is passing through the region.
An electromagnetic wave may be passing through the region.
An electromagnetic wave is certainly passing through the region.

Answer

An electromagnetic wave may be passing through the region.

Explanation:
For an electromagnetic wave,electric field, magnetic field and direction of propagation are mutually perpendicular to each other. We can have a region in which electric and magnetic fields are applied at an angle with each other. In transmission lines Different modes exist. In transverse electric (TE) mode-no electric field exist in the direction of propagation. These are sometimes called H modes because there is only a magnetic field along the direction of propagation (His the conventional symbol for magnetic field).

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Question 101 Mark

The energy contained in a small volume through which an electromagnetic wave is passing oscillates with:

Zero frequency.
The frequency of the wave.
Half the frequency of the wave.
Double the frequency of the wave.

Answer

Double the frequency of the wave.

Explanation:
The energy per unit volume of an electromagnetic wave,
$\text{u}=\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}$
The energy of the given volume can be calculated by multiplying the volume with the above expression.
$\text{U}=\text{u}\times\text{V}=\Big(\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}\Big)\times\text{V}\ ....(\text{i})$
Let the direction of propagation of the electromagnetic wave be along the z-axls. Then, the electric and magnetic fields at a particular point are given by,
$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$
$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$
Substituting the values of electric and magnetic fields in (1) we get,
$\text{U}=\Big(\frac{1}{2}\in_0\big(\text{E}_0^2\sin^2(\text{kz}-\omega\text{t})+\frac{\text{B}^2_0\sin^2(\text{kz}-\omega\text{t})}{2\mu_0}\Big)\times\text{V}$
$\text{U}=\Big(\in_0\text{E}^2_0\frac{(1-\cos(2\text{kz}-2\omega\text{t}))}{4}+\frac{\text{B}_0^2(1-\cos(2\text{kz}-2\omega\text{t}))}{4\mu_0}\Big)\times\text{V}$
From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency $2\omega$ Thus. the frequency of the energy of the electromagnetic wave will also be double.

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Question 111 Mark

Electromagnetic waves are produced by:

A static charge.
A moving charge.
An accelerating charge.
Chargeless particles.

Answer

An accelerating charge.

Explanation:
A static charge produces an electrostatic field. A moving charge produces a magnetic field. Electromagnetic waves are produced by an accelerating charge.

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Question 121 Mark

A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.

$\text{p}=0,\text{ E}\neq0$
$\text{p}\neq0,\text{ E}=0$
$\text{p}\neq0,\text{ E}\neq0$
$\text{p}=0,\text{ E}=0$

Answer

$\text{p}\neq0,\text{ E}\neq0$

Explanation:
When an electromagnetic wave strikes a material surface, it transports the momentum, as well as the energy, to the surface. The striking electromagnetic wave exerts pressure on the surface. The total energy transferred to the surface by the electromagnetic wave is given by $\text{E}=\text{pc}$ Therefore, $\text{p}\neq0,\text{ E}\neq0$

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Question 131 Mark

A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving:

Along the electric field.
Along the magnetic field.
Along the direction of propagation of the wave.
In a plane containing the magnetic field and the direction of propagation.

Answer

Along the electric field.

Explanation:
As the electron is at rest initially, only the electric field will exert force on it. There will be no magnetic force on the electron in the stating. Hence, the electron will start moving along the electric field.

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Question 141 Mark

Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor:

Increases.
Decreases.
Does not change.
Is zero.

Answer

Increases.
Decreases.

Explanation:
Displacement current inside a capacitor,
$\text{i}_\text{d}=\epsilon_0\frac{\text{d}\phi_\text{E}}{\text{dt}},$ where
$\phi_\text{E}$ 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 electric field increases or decreases inside the capacitor.

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Physics M.C.Q (1 Marks)

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