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One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in
(c) Here it is given, the energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms is \(E=11 \mathrm{eV}\)
We know that, \(E=h f\), where \(h=6.62 \times 10^{-34} \mathrm{~J}\)-s
\(
f=\text { frequency }
\)
\(
\begin{aligned}
\Rightarrow \quad 11 \mathrm{eV} & =h f \quad \\
f & =\frac{11 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=2.65 \times 10^{15} \mathrm{~Hz}
\end{aligned}
\)
This frequency radiation belongs to ultraviolet region.
A linearly polarized electromagnetic wave given as \(\mathbf{E}=E_o \hat{\mathbf{i}} \cos (k z-\omega \mathrm{t})\) is incident normally on a perfectly reflecting infinite wall at \(z=a\). Assuming that the material of the wall is optically inactive, the reflected wave will be given as
(b) When a wave is reflected from denser medium, then the type of wave doesn’t change but only its phase changes by \(180^{\circ}\) or \(\pi\) radian.
Thus, for the reflected wave \(\hat{\mathbf{z}}=-\hat{\mathbf{z}}, \hat{\mathbf{i}}=-\hat{\mathbf{i}}\) and additional phase of \(\pi\) in the incident wave.
Given, here the incident electromagnetic wave is,
\(
\mathbf{E}=E_0 \hat{\mathbf{i}} \cos (k z-\omega t)
\)
The reflected electromagnetic wave is given by
\(
\begin{aligned}
\mathbf{E}_r & =E_0(-\hat{\mathbf{i}}) \cos [k(-z)-\omega t+\pi] \\
& =-E_0 \hat{\mathbf{i}} \cos [-(k z+\omega t)+\pi] \\
& =E_0 \hat{\mathbf{i}} \cos \left[-\left(k_z+\omega t\right)=E_0 \hat{\mathbf{i}} \cos (k z+\omega t)\right]
\end{aligned}
\)
Light with an energy flux of \(20 \mathrm{~W} / \mathrm{cm}^2\) falls on a non-reflecting surface at normal incidence. If the surface has an area of \(30 \mathrm{~cm}^2\). the total momentum delivered (for complete absorption) during 30 minutes is
(b) Given, energy flux \(\phi=20 \mathrm{~W} / \mathrm{cm}^2\)
Area, \(\quad A=30 \mathrm{~cm}^2\)
Time, \(t=30 \mathrm{~min}=30 \times 60 \mathrm{~s}\)
Now, total energy falling on the surface in time \(t\) is,
\(
U=\phi A t=20 \times 30 \times(30 \times 60) \mathrm{J}
\)
Momentum of the incident light \(=\frac{U}{c}\)
\(
=\frac{20 \times 30 \times(30 \times 60)}{3 \times 10^8}=36 \times 10^{-4} \mathrm{~kg}-\mathrm{ms}^{-1}
\)
Momentum of the reflected light \(=0\)
\(\therefore\) Momentum delivered to the surface
\(
=36 \times 10^{-4}-0=36 \times 10^{-4} \mathrm{~kg}-\mathrm{ms}^{-1}
\)
As no reflection from the surface and for complete absorption so momentum of reflected radiation is zero.
Momentum delivered to surface \(=\) Change in momentum
\(
=p_f-p_i=0-36 \times 10^{-4}=-36 \times 10^{-4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\)
(-) sign shows the direction of momentum.
The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E . The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is
(c) Electric field intensity on a surface due to incident radiation is,
\(
I_{a v} \propto E_o^2
\)
\(
\begin{aligned}
& \frac{P_{a V .}}{A} \propto E_o^2 \\
& P_{a v} \propto E_o^2 [ [latex]\because A[latex] is same in both cases]
\end{aligned}
\)
\(
\begin{aligned}
&\text { We know that, } E_0 \propto \sqrt{P_{a v}}\\
&\begin{aligned}
& \therefore \frac{\left(E_0\right)_1}{\left(E_0\right)_2}=\sqrt{\frac{\left(P_{a v}\right)_1}{\left(P_{a v}\right)_2}} \\
& \Rightarrow \frac{E}{\left(E_0\right)_2}=\sqrt{\frac{100}{50}} \\
& \left(E_0\right)_2=E / \sqrt{2}
\end{aligned}
\end{aligned}
\)
If \(\mathbf{E}\) and \(\mathbf{B}\) represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along
(d) Key concept: A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. The electric vector is responsible for the optical effects of an EM wave and is called the light vector.
The direction of propagation of electromagnetic wave is perpendicular to both electric field vector \((\vec{E})\) and \(\vec{B}\) magnetic field vector \(B\), i.e., in the direction of \(\vec{E} \times \vec{B}\).
Here, electromagnetic wave is along the \(z\)-direction which is given by the cross product of \(E\) and \(B\).
The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is
Intensity in terms of electric field \(\quad U_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E_o^2\)
Intensity in terms of magnetic field \(U_{\mathrm{av}}=\frac{1}{2} \frac{B_0^2}{\mu_0}\)
Now taking the intensity in terms of electric field.
\(
\left(U_{\mathrm{av}}\right) \text { electric field }=\frac{1}{2} \varepsilon_0 E_o^2
\)
\(
=\frac{1}{2} \varepsilon_0\left(c B_0\right)^2 \quad\left(\because E_0=c B_0\right)
\)
\(
=\frac{1}{2} \varepsilon_0 \times C^2 B^2
\)
\(
\begin{aligned}
&\text { But, }\\
&c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
\end{aligned}
\)
\(
\therefore \quad\left(U_{\mathrm{av}}\right)_{\text {Electric field }}=\frac{1}{2} \varepsilon_0 \times \frac{1}{\mu_0 \varepsilon_0} B_o^2=\frac{1}{2} \frac{B_0^2}{\mu_0}
\)
\(=\left(U_{\mathrm{av}}\right)_{\text {magnetic field }}\)
Thus, the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.
Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is \(1: 1\).
Note:
When the incident EM wave is completely absorbed by a surface, it delivers energy U and momentum \(\mathrm{U} / \mathrm{c}\) to the surface.
When a wave of energy \(U\) is totally reflected from the surface, the momentum delivered to surface is \(2 \mathrm{U} / \mathrm{c}\).
An EM wave radiates outwards from a dipole antenna, with \(E_0\) as the amplitude of its electric field vector. The electric field \(E_0\) which transports significant energy from the source falls off as
(c) From a diode antenna, the electromagnetic waves are radiated outwards.
The amplitude of electric field vector \(\left(E_0\right)\) which transports significant energy from the source falls off intensity inversely as the distance \((r)\) from the antenna, i.e., \(E_0 \propto \frac{1}{r}\).
An electromognetic wave travels in vacuum along \(z\) direction: \(\mathbf{E}=\left(E_1 \hat{\mathbf{i}}+E_2 \hat{\mathbf{j}}\right) \cos (k z-\omega t)\). Choose the correct options from the following:
(a, d) We are given that the electric field vector of an electromagnetic wave travels in a vacuum along z-direction as,
\(
\vec{E}=\left(E_2 \hat{i}+E_2 \hat{j}\right) \cos (k z-\omega t)
\)
The magnitude of the electric and the magnetic fields in an electromagnetic wave are related as
\(
\begin{aligned}
& B_0=\frac{E_0}{c} \\
& \vec{B}=\frac{\vec{E}}{c}=\frac{E_1 \hat{i}+E_2 \hat{j}}{c} \cos (k z-\omega t)
\end{aligned}
\)
Also, \(\vec{E}\) and \(\vec{B}\) are perpendicular to each other and the propagation of electromagnetic wave is perpendicular to \(\vec{E}\) as well as \(\vec{B}\), so the given electromagnetic wave is plane polarized.
An electromagnetic wave travelling along \(z\)-axis is given as: \(\mathbf{E}=\mathbf{E}_{\mathrm{o}} \cos (k z-\omega \mathrm{t}\)) . Choose the correct options from the following;
(a, b, c) The direction of propagation of an electromagnetic wave is always along the direction of vector product \(\vec{E} \times \vec{B}\). Refer to Figure.
\(
\begin{aligned}
\vec{B} & =B \hat{j}=B(\hat{k} \times \hat{i})=\frac{E}{c}(\hat{k} \times \hat{i}) \\
& =\frac{1}{c}[k \times E \hat{i}]=\frac{1}{c}[\hat{k} \times \vec{E}] \quad\left(\text { as } \frac{E}{B}=c\right)
\end{aligned}
\)
(b) \(\vec{E}=E \hat{i}=c B(\hat{j} \times \hat{k})=c(B \hat{j} \times \hat{k})=c(\vec{B} \times \hat{k})\)
(c) \(\hat{k} \cdot \vec{E}=\hat{k} \cdot(E \hat{i})=0, \vec{k} \cdot \vec{B}=\hat{k} \cdot(B \hat{j})=0\)
(d)
\(
\begin{aligned}
& \hat{k} \times \vec{E}=\hat{k} \times(E \hat{i})=E(\hat{k} \times \hat{i})=E \hat{j} \\
& \text { and } \hat{k} \times \vec{B}=\hat{k} \times(B \hat{j})=B(\hat{k} \times \hat{j})=-B \hat{i}
\end{aligned}
\)
A plane electromagnetic wave propagating along \(x\) direction can have the following pairs of \(\mathbf{E}\) and \(\mathbf{B}\)
(b, d) As electric and magnetic field vectors \(\mathbf{E}\) and \(\mathbf{B}\) are perpendicular to each other as well as perpendicular to the direction of propagation of electromagnetic wave.
Here in the question electromagnetic wave is propagating along \(x\)-direction. So, electric and magnetic field vectors should have either \(y\)-direction or \(z\)-direction.
A charged particle oscillates about its mean equilibrium position with a frequency of \(10^9 \mathrm{~Hz}\). The electromagnetic waves produced:
(a, c, d) Here we are given the frequency by which the charged particles oscillates about its mean equilibrium position, it is equal to \(10^9 \mathrm{~Hz}\). The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.
So, frequency of electromagnetic waves produced by the charged particle is \(\nu=10^9 \mathrm{~Hz}\).
Wavelength \(\lambda=\frac{c}{\nu}=\frac{3 \times 10^8}{10^9}=0.3 \mathrm{~m}\)
The frequency of \(10^9 \mathrm{~Hz}\) falls in the region of radiowaves.
The source of electromagnetic waves can be a charge
(b, d) Key concept:
An electromagnetic wave can be produced by accelerated or oscillating charge.
An oscillating charge is accelerating continuously, it will radiate electromagnetic waves continuously.
Electromagnetic waves are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number.
Here, in option (b) charge is moving in a circular orbit.
In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.
In option (d), the charge is falling in electric field. If a charged particle is moving in electric field it experiences a force or we can say it accelerates. We know an accelerating charge particle radiates electromagnetic waves. Hence option (d) is also correct.
Note: Also, we know that a charge starts accelerating when it falls in an electric field. Important points:
In an atom an electron is circulating around the nucleus in a stable orbit, although accelerating does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit.
A simple LC oscillator and energy source can produce waves of desired frequency.
An EM wave of intensity \(I\) falls on a surface kept in vacuum and exerts radiation pressure \(p\) on it. Which of the following are true?
(a, c, d) Radiation pressure \((p)\) is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.
Momentum per unit time per unit area
\(
=\frac{\text { Intensity }}{\text { Speed of wave }}=\frac{I}{c}
\)
Change in momentum per unit time per unit area \(=\frac{\Delta I}{c}=\) radiation pressure \((p)\)
i.e., \(p=\frac{\Delta I}{c}\)
Momentum of incident wave per unit time per unit area \(=\frac{I}{\mathrm{C}}\)
When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area \(=0\).
Radiation pressure \((p)=\) change in momentum per unit time per unit area \(=\frac{\Delta I}{\mathrm{c}}=\frac{I}{\mathrm{c}}-0=\frac{I}{\mathrm{c}}\).
When wave is totally reflected, then momentum of the reflected wave per unit time per unit area \(=-\frac{I}{\mathrm{C}}\), Radiation pressure \(p=\frac{I}{\mathrm{c}}-\left(-\frac{I}{\mathrm{c}}\right)=\frac{2 I}{\mathrm{c}}\).
Here, \(p\) lies between \(\frac{I}{c}\) and \(\frac{2 I}{c}\).
A magnetic field can be produced by
(d)Â A magnetic field can be produced by both a moving charge and a changing electric field.
Explanation:
According to Maxwell’s equations, a changing electric field generates a magnetic field, and a moving charge also creates a magnetic field. This relationship between electricity and magnetism is fundamental in understanding electromagnetic waves.
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
(d) 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 \(t\) is given by :
\(
Q=C V\left(1-e^{-\mathrm{t} / \mathrm{RC}}\right),
\)
where
\(Q=\) charge developed on the plates of the capacitor
\(R=\) resistance of the resistor connected in series with the capacitor
\(\mathrm{C}=\) capacitance of the capacitor
\(\mathrm{V}=\) potential difference of the battery
The time constant of the capacitor is given, \(\tau=\mathrm{RC}\)
The capacitor keeps on charging up to the time \(\tau\). The development of charge on the plates will be gradual after \(\mathrm{t}=\mathrm{RC}\). 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.
Note:
Deflects and gradually comes to the original position in a time which is large compared to the time constant.
Explanation:
Charging process:
When the capacitor is connected to a battery, it starts charging, creating an electric field between the plates.
Magnetic field generation:
This changing electric field induces a magnetic field around the capacitor, which can affect a nearby compass needle, causing it to deflect.
Time constant:
The rate of charging is determined by the time constant of the circuit, which is equal to the product of the capacitance and resistance.
Gradual return:
As the capacitor charges, the electric field and magnetic field stabilize, causing the compass needle to gradually return to its original position, but this process takes a time significantly longer than the time constant.
Dimensions of \(1 /\left(\mu_0 \varepsilon_0\right)\) is
(c) We know that \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\).
Squaring both sides gives \(c^2=\left(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right)^2\). This simplifies to \(c^2=\frac{1}{\mu_0 \varepsilon_0}\).
The dimensions of speed, \(c[latex], are [latex]\frac{\mathrm{L}}{\mathrm{T}}\).
Therefore, the dimensions of \(c^2\) are \(\left(\frac{\mathrm{L}}{\mathrm{T}}\right)^2\).
\(
\text { The dimensions of } \frac{1}{\mu_0 \varepsilon_0} \text { are } \frac{\mathrm{L}^2}{\mathrm{~T}^2} \text {. }
\)
Electromagnetic waves are produced by
(c) An accelerating charge.
Explanation: When a charged particle accelerates, it creates changing electric and magnetic fields that propagate through space as electromagnetic waves. These waves can then travel at the speed of light and carry energy.
Why other options are incorrect:
(a) a static charge:
A static charge only creates an electric field around itself. It doesn’t produce a wave.
(b) a moving charge:
While a moving charge produces both electric and magnetic fields, it only creates an electromagnetic wave if the motion is accelerating. A charge moving at a constant velocity does not produce electromagnetic waves.
(d) chargeless particles:
Particles without a charge cannot produce electric or magnetic fields, and therefore cannot produce electromagnetic waves.
An electromagnetic wave going through vacuum is described by
\(
E=E_0 \sin (k x-\omega t) ; B=B_0 \sin (k x-\omega t)
\)
Then
(a) The speed of light \(c\) in vacuum is given by the relationship:
\(
c=\frac{E_0}{B_0}
\)
We also know that the wave speed can be expressed in terms of wave number \(k\) and angular frequency \(\omega\) :
\(
c=\frac{\omega}{k}
\)
Set the two expressions for \(c\) equal to each other:
Since both expressions represent the speed of light, we can equate them:
\(
\frac{E_0}{B_0}=\frac{\omega}{k}
\)
Cross-multiply to find a relationship between \(E_0, B_0, k\), and \(\omega\) :
Rearranging the equation gives:
\(
E_0 k=B_0 \omega
\)
An electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) exist in a region. The fields are not perpendicular to each other.
(c) If an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) exist in a region and are not perpendicular to each other, it is possible that an electromagnetic wave is passing through that region.
Explanation:
While electromagnetic waves are characterized by mutually perpendicular electric and magnetic fields, this is not the only condition for a region to contain both fields. An electromagnetic wave is a specific type of wave where the electric and magnetic fields are orthogonal (perpendicular) to each other and to the direction of propagation. However, other configurations of electric and magnetic fields can exist in a region, where they are not perpendicular, but still carry electromagnetic energy.
Why other options are incorrect:
(a) This is not possible:
This is incorrect. It is possible to have an electric and magnetic field in the same region even when they are not perpendicular.
(b) No electromagnetic wave is passing through the region:
This is too strong of a statement. While the fields might not be perpendicular, it’s not impossible for them to also be part of an electromagnetic wave. The presence of nonperpendicular fields does not preclude the presence of a perpendicular component which would constitute an electromagnetic wave.
(d) An electromagnetic wave is certainly passing through the region:
This is also incorrect. While electromagnetic waves can propagate in regions where the fields are not perpendicular, it is not guaranteed. The perpendicularity of the fields is a characteristic of the wave itself, not just the presence of the fields.
Consider the following two statements regarding a linearly polarized, plane electromagnetic wave:
(A) The electric field and the magnetic field have equal average values.
(B) The electric energy and the magnetic energy have equal average values.
(a) For a linearly polarised, plane electromagnetic wave
\(
\begin{aligned}
& E=E_0 \sin \omega(t-x / c) \\
& B=B_0 \sin \omega(t-x / c)
\end{aligned}
\)
The average value of either \(E\) or \(B\) over a cycle is zero ( average of \(\sin (\theta)\) over a cycle is zero).
Also the electric energy density \(\left(u_E\right)\) and magnetic energy density \(\left(u_B\right)\) are equal.
\(
u_E=\frac{1}{2} \in_0 E^2=\frac{B^2}{2 u_0}=u_B
\)
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.
A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving
(a) Electromagnetic wave contains both electric and magnetic field. When electron is placed, it experience force due to electric field because of which it start moving along the electric field. Force due to magnetic field is zero.
Note: The magnetic field component of the wave does not directly cause the electron to move in its direction. The force due to the magnetic field is perpendicular to both the magnetic field and the electron’s velocity, according to the Lorentz force law, according to physics. Since the electron starts from rest, this force is initially zero.
A plane electromagnetic wave is incident on a material surface. The wave delivers momentum \(p\) and energy \(E\).
(c) Electromagnetic waves are composed of oscillating electric and magnetic fields. These fields carry both energy and momentum.
When an electromagnetic wave strikes a material surface, it interacts with the charges within the material.
This interaction results in the transfer of both energy and momentum to the surface.
Since both energy and momentum are transferred, neither \(p\) nor \(E\) can be zero. Therefore, \(p \neq 0\) and \(E \neq 0\).
When a plane electromagnetic wave is incident on a material surface, it delivers both momentum and energy, so \(p \neq 0\) and \(E \neq 0\).
An electromagnetic wave going through vacuum is described by
\(
E=E_0 \sin (k x-\omega t)
\)
Which of the following is/are independent of the wavelength?
(c) The angular wave number \(k=\frac{2 \pi}{\lambda}\); where \(\lambda\) is the wavelength.
The angular frequency is \(\omega=2 \pi f\).
The ratio \(\frac{k}{\omega}=\frac{2 \pi / \lambda}{2 \pi f}\)
\(
=\frac{1}{f \lambda}=\frac{1}{c}=\text { constant }
\)
Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor
(a, b) Displacement current inside a capacitor, \(i_d=\epsilon_0 \frac{d \Phi_E}{d t}\), 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 electric field increases or decreases inside the capacitor.
Explanation: Displacement current flows through the gap between capacitor plates when the charge on the plates is changing, which means either increasing or decreasing.
Displacement current and changing electric field:
Displacement current, as described by Maxwell’s equations, arises when there is a changing electric field or flux within a region. In a capacitor, this happens when the charge on the plates is changing, causing the electric field between the plates to change according to physics resources.
Charging and discharging:
During the charging process, the charge on the capacitor plates increases, creating a displacement current. Similarly, during discharging, the charge decreases, and a displacement current also exists, as the electric field is changing.
No displacement current when constant:
When the capacitor is fully charged (or completely discharged), the charge on the plates and the electric field between them remain constant, and there is no displacement current, according to an online physics resource.
Speed of electromagnetic waves is the same
Electromagnetic waves (EM waves) travel at a speed denoted by ‘ \(c\) ‘ in a vacuum, which is approximately \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\). This speed can change when EM waves travel through different media.
Explanation:
The speed of electromagnetic waves is influenced by the medium through which they travel. In a vacuum, all EM waves travel at the same speed. However, when they pass through a medium, their speed can vary based on the medium’s refractive index. The speed of an electromagnetic wave is related to its frequency (f) and wavelength ( \(\lambda\) ) by the equation:
\(
c=f \cdot \lambda
\)
If the frequency changes, the wavelength must also change to keep the speed constant in a vacuum. However, in a medium, different frequencies can result in different speeds due to varying refractive indices.
Which of the following have zero average value in a plane electromagnetic wave?
(a, b) In a plane electromagnetic wave, both the (a) electric field and (b) magnetic field have zero average value.
Explanation: In a plane electromagnetic wave, the electric and magnetic fields oscillate sinusoidally, and over a complete cycle, the average value of a sine wave is zero.
Why other options are incorrect:
(c) Electric energy:
While the electric field itself has a zero average value, the electric energy is proportional to the square of the electric field. Since the square of a sinusoidal function has a positive average value, the electric energy also has a nonzero average value.
(d) Magnetic energy:
Similar to electric energy, magnetic energy is proportional to the square of the magnetic field and therefore also has a nonzero average value.
The energy contained in a small volume through which an electromagnetic wave is passing oscillates with
(d) The energy density of an electromagnetic wave is given by the formula:
\(
U=\frac{1}{2} \epsilon_0 E^2+\frac{1}{2 \mu_0} B^2
\)
where \(U\) is the energy density, \(E\) is the electric field, \(B\) is the magnetic field, \(\epsilon_0\) is the permittivity of free space, and \(\mu_0\) is the permeability of free space.
Write the expressions for electric and magnetic fields:
The electric field \(E\) can be expressed as:
\(
E=E_0 \sin (k z-\omega t)
\)
The magnetic field \(B\) can be expressed as:
\(
B=B_0 \sin (k z-\omega t)
\)
Substitute the fields into the energy density formula:
Substitute \(E\) and \(B\) into the energy density equation:
\(
\begin{aligned}
U & =\frac{1}{2} \epsilon_0\left(E_0 \sin (k z-\omega t)\right)^2 \\
+ & \frac{1}{2 \mu_0}\left(B_0 \sin (k z-\omega t)\right)^2
\end{aligned}
\)
\(
U=\frac{1}{2} \epsilon_0 E_0^2 \sin ^2(k z-\omega t)+\frac{1}{2 \mu_0} B_0^2 \sin ^2(k z-\omega t)
\)
\(
\begin{aligned}
& \text { Recall that } \sin ^2(x)=\frac{1-\cos (2 x)}{2}: \\
& U=\frac{1}{2} \epsilon_0 E_0^2\left(\frac{1-\cos (2(k z-\omega t))}{2}\right) \\
& +\frac{1}{2 \mu_0} B_0^2\left(\frac{1-\cos (2(k z-\omega t))}{2}\right)
\end{aligned}
\)
\(
\begin{aligned}
&\text { This can be simplified to: }\\
&\begin{gathered}
U=\left(\frac{\epsilon_0 E_0^2}{4}+\frac{B_0^2}{4 \mu_0}\right) \\
-\left(\frac{\epsilon_0 E_0^2+B_0^2}{4} \cos (2(k z-\omega t))\right)
\end{gathered}
\end{aligned}
\)
The term \(\cos (2(k z-\omega t))\) indicates that the energy oscillates at a frequency of \(2 \omega\), which is double the frequency of the electromagnetic wave.
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 A/2 parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is:
(a) Suppose the charge on the capacitor at time \(t\) is \(Q\). The electric field between the plates of the capacitor is \(E=\frac{Q}{\varepsilon_0 A}\). The flux through the area considered is
\(
\Phi_E=\frac{Q}{\varepsilon_0 A} \cdot \frac{A}{2}=\frac{Q}{2 \varepsilon_0}
\)
The displacement current is
\(
i_d=\varepsilon_0 \frac{d \Phi_E}{d t}=\varepsilon_0\left(\frac{1}{2 \varepsilon_0}\right) \frac{d Q}{d t}=\frac{i}{2}
\)
A plane electromagnetic wave propagating in the \(x\)-direction has a wavelength of 5.0 mm . The electric field is in the \(y\)-direction and its maximum magnitude is \(30 \mathrm{~V} \mathrm{~m}^{-1}\). Equations for the electric and magnetic fields as a function of \(x\) and \(t\) is:
(a) The equation for the electric and the magnetic fields in the wave may be written as
\(
\begin{aligned}
& E=E_0 \sin \omega\left(t-\frac{x}{c}\right) \\
& B=B_0 \sin \omega\left(t-\frac{x}{c}\right)
\end{aligned}
\)
We have,
\(
\omega=2 \pi v=\frac{2 \pi}{\lambda} c .
\)
\(
\begin{aligned}
&\text { Thus, }\\
&\begin{aligned}
E & =E_0 \sin \left[\frac{2 \pi}{\lambda}(c t-x)\right] \\
& =\left(30 \mathrm{~V} \mathrm{~m}^{-1}\right) \sin \left[\frac{2 \pi}{5 \cdot 0 \mathrm{~mm}}(c t-x)\right] .
\end{aligned}
\end{aligned}
\)
The maximum magnetic field is
\(
B_0=\frac{E_0}{c}=\frac{30 \mathrm{~V} \mathrm{~m}^{-1}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=10^{-7} \mathrm{~T} .
\)
\(
\begin{aligned}
&\text { So, }\\
&\begin{aligned}
B & =B_0 \sin \left[\frac{2 \pi}{\lambda}(c t-x)\right] \\
& =\left(10^{-7} \mathrm{~T}\right) \sin \left[\frac{2 \pi}{5 \cdot 0 \mathrm{~mm}}(c t-x)\right] .
\end{aligned}
\end{aligned}
\)
The magnetic field is along the \(z\)-axis.
A light beam travelling in the \(x\)-direction is described by the electric field \(E_y=\left(300 \mathrm{~V} \mathrm{~m}^{-1}\right) \sin \omega(t-x / c)\). An electron is constrained to move along the \(y\)-direction with a speed of \(2.0 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\). The maximum electric force and the maximum magnetic force on the electron is:
(c) The maximum electric field is \(E_0=300 \mathrm{~V} \mathrm{~m}^{-1}\). The maximum magnetic field is
\(
B_0=\frac{E_0}{c}=\frac{300 \mathrm{~V} \mathrm{~m}^{-1}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=10^{-6} \mathrm{~T}
\)
along the \(z\)-direction.
The maximum electric force on the electron is
\(
\begin{aligned}
F_e & =q E_0=\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(300 \mathrm{~V} \mathrm{~m}^{-1}\right) \\
& =4.8 \times 10^{-17} \mathrm{~N} .
\end{aligned}
\)
The maximum magnetic force on the electron is
\(
F_b=|q \vec{v} \times \vec{B}|_{\max }=q v B_0
\)
\(
\begin{aligned}
& =\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(2.0 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\right) \times\left(10^{-6} \mathrm{~T}\right) \\
& =3.2 \times 10^{-18} \mathrm{~N} .
\end{aligned}
\)
The energy stored in a 60 cm length of a laser beam operating at 4 mW is:
(d) The time taken by the electromagnetic wave to move through a distance of 60 cm is \(t=\frac{60 \mathrm{~cm}}{c}=2 \times 10^{-9} \mathrm{~s}\). The energy contained in the 60 cm length passes through a cross-section of the beam in \(2 \times 10^{-9} \mathrm{~s}\). But the energy passing through any cross section in \(2 \times 10^{-9} \mathrm{~s}\) is
\(
U=(4 \mathrm{~mW}) \times\left(2 \times 10^{-9} \mathrm{~s}\right)
\)
\(
\begin{aligned}
&\begin{aligned}
& =\left(4 \times 10^{-3} \mathrm{Js}^{-1}\right) \times\left(2 \times 10^{-9} \mathrm{~s}\right) \\
& =8 \times 10^{-12} \mathrm{~J}
\end{aligned}\\
&\text { This is the energy contained in } 60 \mathrm{~cm} \text { length. }
\end{aligned}
\)
The amplitude of the electric field in a parallel beam of light of intensity \(2.0 \mathrm{Wm}^{-2}\) is
(b) The intensity of a plane electromagnetic wave is
\(
I=u_{a v} c=\frac{1}{2} \varepsilon_0 E_0^2 c
\)
\(E_0=\sqrt{\frac{2 I}{\varepsilon_0 c}}\)
\(
=\sqrt{\frac{2 \times\left(2.0 \mathrm{~W} \mathrm{~m}^{-2}\right)}{\left(8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}}
\)
According to Maxwell’s hypothesis, a changing electric field gives rise to
(c)Â According to Maxwell’s hypothesis, a changing electric field gives rise to a magnetic field.
Explanation: Maxwell’s equations, particularly the Ampère-Maxwell law, state that a changing electric field creates a magnetic field. This principle is fundamental to the propagation of electromagnetic waves, where the changing electric field induces a changing magnetic field, and vice versa.
Maxwell in his famous equation of electromagnetism introduced the concept of
(c) Displacement current. Maxwell introduced the concept of displacement current in his equations of electromagnetism to explain how changing electric fields can produce magnetic fields, even in the absence of a traditional electric current.
Explanation:
Displacement Current:
Maxwell’s equations, which unified electricity and magnetism, include a term representing the displacement current, which is essentially a time-varying electric field. This term was crucial for explaining phenomena like the propagation of electromagnetic waves.
\(
I_d=\varepsilon_0 \frac{d \phi_E}{d t}
\)
Dimensions of \(\varepsilon_0 \frac{d \phi_E}{d t}\) are same as that of
(d) We have, displacement current, \(I_d=\varepsilon_0 \frac{d \phi_E}{d t}\)
So, dimensions of \(\varepsilon_0 \frac{d \phi_E}{d t}\) are same as that of current.
According to Ampere-Maxwell’s circuital law,
\(
\begin{aligned}
&\text { (d) Ampere-Maxwell’s circuital law is given by }\\
&\oint \mathbf{B} \cdot d \mathbf{l}=\mu_0 \cdot\left(I_c+I_d\right)=\mu_0\left(I_c+\varepsilon_0 \frac{d \phi_E}{d t}\right)
\end{aligned}
\)
The conduction current in ideal case through a circuit is zero when charge on capacitor is
(b) When capacitor is fully charged (i.e. charge on plates of capacitor is maximum) no current flows through the circuit. The potential of the plates became constant and there is no change in the electric field between the plates. So, both conduction current in the circuit and displacement current between the plates became zero.
What is the displacement current between the square plate of side 1 cm of a capacitor, if electric field between the plates is changing at the rate of \(3 \times 10^6 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\)?
(a) We know that, displacement current, \(I_d=\varepsilon_0 A \frac{d E}{d t}\)
We have, \(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
Area, \(A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\)
and \(\frac{d E}{d t}=3 \times 10^6 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\)
\(\therefore \quad I_d=8.85 \times 10^{-12} \times 10^{-4} \times 3 \times 10^6 \simeq 2.7 \times 10^{-9} \mathrm{~A}\)
A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. What is the value of the displacement current for \(10^{-6} \mathrm{~s}\), if the plate area is \(60 \mathrm{~cm}^2\)?
\(
\text { (a) Displacement current, } I_d=\varepsilon_0 \frac{d \phi_E}{d t}
\)
\(
\begin{aligned}
& =\varepsilon_0 \frac{E A}{t}=\frac{\varepsilon_0(V / d) \times A}{t}=\frac{\varepsilon_0 V A}{t d} \\
& =\frac{8.85 \times 10^{-12} \times 400 \times\left(60 \times 10^{-4}\right)}{10^{-6} \times\left(2 \times 10^{-3}\right)} \\
& =1.062 \times 10^{-2} \mathrm{~A}
\end{aligned}
\)
An \(L-C\) resonant circuit contains a 200 pF capacitor and a \(100 \mu \mathrm{H}\) inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is
(b) Frequency of \(L\)-C oscillations,
\(\begin{aligned} f & =\frac{1}{2 \pi \sqrt{L C}} \\ & =\frac{1}{2 \times 3.14 \sqrt{100 \times 10^{-6} \times 200 \times 10^{-12}}} \\ & =1.126 \times 10^6 \mathrm{~Hz}\end{aligned}\)
So, wavelength,
\(
\begin{aligned}
\lambda=\frac{c}{v} & =\frac{3 \times 10^8}{1.126 \times 10^6} \\
& =266.4 \simeq 266 \mathrm{~m}
\end{aligned}
\)
The velocity of electromagnetic wave is parallel to
(b) According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vector at right angles to each other as well as the right angles to the direction of wave propagation.
If the electric field \((\vec{E})\) and magnetic field \((\vec{B})\) are vibrating aling
Y and Z direction, the propagation of electromagnetic wave will be along the X-axis. Therefore, the velocity of electromagnetic wave is parallel to \(\vec{E} \times \vec{B}\).
Which of the following statement is incorrect about electromagnetic waves?
(c) “They travel with the same speed in all media”.
Explanation: Electromagnetic waves travel at the same speed (the speed of light) in a vacuum, but their speed can vary depending on the medium they are traveling through.
Why other options are correct:
(a) These are transverse in nature: Electromagnetic waves are transverse waves, meaning their electric and magnetic fields oscillate perpendicular to the direction of wave propagation.
(b) These are produced by accelerating charges: When charged particles accelerate, they produce electromagnetic waves.
(d) They travel in free space with the speed of light: In a vacuum (free space), electromagnetic waves travel at the speed of light (approximately \(3 \times 10^{\wedge} 8\) \(\mathrm{m} / \mathrm{s}\) ).
In electromagnetic waves, the phase difference between electric and magnetic field vectors \(\mathbf{E}\) and \(\mathbf{B}\) is
(a) Electromagnetic waves consist of oscillating electric and magnetic fields. These fields are perpendicular to each other and to the direction of wave propagation.
In an electromagnetic wave, the electric field \(\mathbf{E}\) and magnetic field \(\mathbf{B}\) oscillate in phase.
This means their peaks and troughs occur simultaneously.
Therefore, the phase difference between them is zero.
The oscillating electric and magnetic field vectors of an electromagnetic waves are oriented along
(c) Mutually perpendicular directions and are in phase. In electromagnetic waves, the electric and magnetic field vectors are always perpendicular to each other and also perpendicular to the direction of wave propagation. Furthermore, they oscillate in phase, meaning their crests and troughs align simultaneously.
Amongst the following relations, which is correct?
(b) As, \(\frac{E_0}{B_0}=c=\frac{1}{\sqrt{\varepsilon_0 \mu_0}} \Rightarrow \sqrt{\mu_0 \varepsilon_0}=\frac{B_0}{E_0}\)
Here, \(\quad c=\) speed of light
In an apparatus, the electric field was found to oscillate with amplitude of \(18 \mathrm{Vm}^{-1}\). The magnitude of the oscillating magnetic field will be
\(
\text { (b) As, } c=\frac{E_0}{B_0} \Rightarrow B_0=\frac{E_0}{c}=\frac{18}{3 \times 10^8}=6 \times 10^{-8} \mathrm{~T}
\)
The rms value of the electric field of the light coming from the sun is \(720 \mathrm{NC}^{-1}\). The average total energy density of the electromagnetic wave is
\(
\begin{aligned}
&\text { (d) Average energy density, }\\
&\begin{aligned}
u_{\mathrm{av}} & =\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \varepsilon_0\left(\sqrt{2} E_{\mathrm{rms}}\right)^2=\varepsilon_0 E_{\mathrm{rms}}^2 \\
& =8.85 \times 10^{-12} \times(720)^2=4.58 \times 10^{-6} \mathrm{Jm}^{-3}
\end{aligned}
\end{aligned}
\)
The electric field of a plane electromagnetic wave varies with time of amplitude \(2 \mathrm{Vm}^{-1}\) propagating along \(Z\)-axis. The average density of the magnetic field is (in \(\mathrm{J} \mathrm{m}^{-3}\) )
(c) Amplitude of electric field and magnetic field are related by the relation, \(\frac{E_0}{B_0}=c\)
Average energy density of the magnetic field,
\(
u_B=\frac{1}{2} \frac{B_0^2}{\mu_0}=\frac{1}{2} \frac{E_0^2}{\mu_0 c^2} \quad\left(\because B_0=\frac{E_0}{c}\right)
\)
\(
=\frac{1}{2} \varepsilon_0 E_0^2 \quad\left(\because c=\frac{1}{\sqrt{\mu_0 E_0}}\right)
\)
\(
\begin{aligned}
& =\frac{1}{2} \times 8.85 \times 10^{-12} \times 2^2 \\
& =17.71 \times 10^{-12} \mathrm{Jm}^{-3}
\end{aligned}
\)
In an electromagnetic wave, the electric and magnetic fields are \(100 \mathrm{Vm}^{-1}\) and 0.265 T , respectively. The maximum rate of energy flow is
(a) Here, \(E_0=100 \mathrm{Vm}^{-1}, B_0=0.265 \mathrm{~T}\)
\(\therefore\) Maximum rate of energy flow, \(S=\frac{E_0 \times B_0}{\mu_0}=\frac{100 \times 0.265}{4 \pi \times 10^{-7}}\)
\(
=21 \times 10^6 \mathrm{Wm}^{-2}
\)
Radiations of intensity \(0.5 \mathrm{Wm}^{-2}\) are striking a metal plate. The pressure on the plate is
(b) As metal is reflecting surface and for this type of surface radiation pressure is given as
\(
p_r=\frac{2 S}{c}=\frac{2 \times 0.5}{3 \times 10^8}=0.33 \times 10^{-8} \mathrm{Nm}^{-2}
\)
Which of the following waves have the maximum wavelength?
(d) Radio waves.
Explanation: Radio waves have the longest wavelengths among the options listed.
Why other options are incorrect:
(a) X-rays: X-rays have much shorter wavelengths than radio waves.
(b) Infrared waves: Infrared waves have wavelengths longer than visible light but significantly shorter than radio waves.
(c) Ultraviolet: Ultraviolet waves are even shorter in wavelength than visible light, and much shorter than radio waves.
Which of the following electromagnetic waves are used in weather forecasting?
(a) The electromagnetic waves used in weather forecasting are Infrared waves.
Explanation: Infrared waves are used to detect temperature variations on the Earth’s surface and in the atmosphere, which is crucial for weather prediction.
Why other options are incorrect:
(b) Radio waves:
While radio waves are used in weather forecasting, they are primarily used in radar systems to detect precipitation and track storm movement.
(c) Microwaves:
Similar to radio waves, microwaves are used in radar systems for weather forecasting.
(d) Visible rays:
Visible light is important for observing cloud cover and other visual conditions, but it doesn’t provide the temperature information necessary for detailed weather forecasting.
Electromagnetic radiation of highest frequency is
(d) \(\gamma\)-rays.
Explanation:
Gamma rays:
These have the highest frequency and shortest wavelength within the electromagnetic spectrum, meaning they carry the most energy.
Infrared radiation:
While considered higher in frequency than radio waves and visible light, infrared radiation still has a significantly lower frequency compared to gamma rays.
Visible radiation:
Visible light falls between infrared and ultraviolet radiation on the spectrum, with a moderate frequency range.
Radio waves:
Radio waves have the lowest frequency and longest wavelength within the electromagnetic spectrum.
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