CBSE Class 12 Physics Electromagnetic Waves Boards Questions Worksheet

This unit introduces the concept of electromagnetic waves, focusing on their generation, propagation, and characteristics. Key topics include Chapter 8 on Electromagnetic Waves, which covers the displacement current hypothesis, the transverse and self-propagating nature of electric and magnetic fields, and the complete electromagnetic spectrum (from radio waves to gamma rays) along with their applications and properties.

Question 501. What is meant by displacement current ? 
Answer: Displacement current is defined as the current that arises in a region of space where the electric field (and hence the electric flux) is changing with time. It is mathematically expressed as: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \] where \( \Phi_E \) is the electric flux and \( \varepsilon_0 \) is the permittivity of free space.
In simple words: Displacement current is not a flow of real electrical charges, but a virtual current produced whenever an electric field changes over time.

Exam Tip: Write down the formula \( I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \) to get full marks on definition questions.

Question 502. In which situation there is a displacement current but no conduction current ? 
Answer: This situation occurs in the empty space between the plates of a capacitor while it is undergoing the process of charging or discharging, or more generally, in any spatial region containing a time-varying electric field with no physical charge carriers.
In simple words: This happens inside the empty gap of a capacitor while it is charging or discharging. Since no real charges can cross the gap, there is only a changing electric field, which acts as a displacement current.

Exam Tip: Specify "between the plates of a charging capacitor" as it is the most common and accepted physical example.

Question 503. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates ?
Answer: During the charging of a capacitor, the displacement current \( I_D \) established in the space between the plates is always equal to the conduction current \( I_C \) flowing through the connecting wires. Therefore, the displacement current across the plates is: \[ I_D = 0.25\text{ A} \]
In simple words: The virtual current inside the capacitor gap is always exactly equal to the real current flowing in the wires, which is 0.25 Amperes.

Exam Tip: Clearly state the property \( I_D = I_C \) before writing down the final numerical value.

Question 504. Why is the quantity \( \varepsilon_0 \frac{d\Phi_E}{dt} \) is called displacement current ? 
Answer: The term \( \varepsilon_0 \frac{d\Phi_E}{dt} \) is called displacement current because it has the same physical dimensions as electric current (Ampere) and produces a magnetic field just like a regular conduction current does when charge displacement or changing electric fields occur in a region, such as between charging capacitor plates.
In simple words: This term is called a current because it has the exact same unit (Amperes) and produces the exact same magnetic field effect as a real current flowing through a wire.

Exam Tip: Mention that it shares the same dimensional formula and magnetic effect as conduction current to make your explanation complete.

Question 505. How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for displacement current in terms of the rate of change of electric flux.
Answer: During the charging process, the accumulating charge on the capacitor plates causes a continuously changing electric field in the gap between them. This changing field produces a time-varying electric flux, which generates a displacement current \( I_D \) inside the gap. The Ampere-Maxwell law incorporates this effect to show that the current path remains unbroken: conduction current flows in the wires, and displacement current flows across the plate gap.
The expression for displacement current in terms of electric flux is: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \]
In simple words: As the capacitor charges, the changing voltage creates a changing electric field in the empty gap. This changing field acts as a virtual "displacement current" that bridges the gap, keeping the circuit's current loop continuous.

Exam Tip: State how the conduction current in the wire seamlessly transitions to displacement current in the gap to explain the continuity of current.

Question 506. Why does a galvanometer show a momentary deflection, at the time of charged or discharging a capacitor ? Write the necessary expression to explain this observation. 
Answer: While a capacitor is actively charging or discharging, the electric field between its plates changes with time, establishing a displacement current \( I_D \) across the gap. This displacement current completes the electrical circuit loop, allowing conduction current to flow through the wires and cause a momentary deflection in the connected galvanometer. Once the capacitor is fully charged or discharged, the electric field stops changing, the displacement current drops to zero, and the deflection vanishes.
The governing expression for this displacement current is: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \]
In simple words: The galvanometer only twitches while the capacitor is charging or discharging because the changing electric field creates a temporary displacement current that completes the circuit. Once charging stops, the field stops changing and the current disappears.

Exam Tip: Note that the deflection is only "momentary" because displacement current only exists when the electric field is actively changing.

Question 507. A capacitor has been charged by a d.c. source. What are the magnitudes of conduction and displacement currents, when it is fully charged ?
Answer: Once the capacitor becomes fully charged by a DC source, the voltage across its plates equals the source voltage, and charge accumulation stops. Because there is no further movement of charges or change in the electric field:
* The magnitude of conduction current is \( I_C = 0 \).
* The magnitude of displacement current is \( I_D = 0 \).
During the active charging phase, both currents are equal and non-zero: \[ I_C = I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \]
In simple words: When fully charged, everything stops. No more charge flows in the wires (conduction current is zero), and the electric field in the gap stops changing (displacement current is zero).

Exam Tip: Be sure to specify that *both* currents become zero at full charge, but were equal and non-zero during charging.

Question 508. What does the displacement current ID = \( \varepsilon_0 \frac{d\Phi_E}{dt} \) signify ? 
Answer: The displacement current signifies that a time-varying electric field (which causes a changing electric flux) can generate a magnetic field in its surrounding space, just as a physical current of moving charges does. This bridges the conceptual link between electricity and magnetism.
In simple words: It proves that a changing electric field doesn't just exist in isolation - it actually produces a magnetic field, just like a real current flowing through a wire.

Exam Tip: Mention "production of a magnetic field by a changing electric field" as the core physical significance.

Question 509. When an ideal capacitor is charged by a d.c. battery, no current flows. However, when an a.c. source is used, the current flows continuously. How does one explain this, based on the concept of displacement current ? 
Answer: The behaviors are described below:
1. **With a DC Source:** After a very brief transient charging phase, the voltage across the plates becomes constant. Since the electric field between the plates stops changing, the rate of change of electric flux is zero, which means the displacement current is zero (\( I_D = 0 \)). The circuit remains broken, and no current flows.
2. **With an AC Source:** The applied voltage alternates continuously, which causes the electric field and the electric flux between the plates to change constantly. This continuous variation generates a steady displacement current (\( I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \)) in the gap, completing the circuit and allowing current to flow continuously.
In simple words: A DC source charges the capacitor once and then stops, so the field stops changing and no more current can flow. An AC source constantly changes direction, keeping the electric field in the gap vibrating forever, which maintains a continuous displacement current.

Exam Tip: Structure your answer into two distinct, labeled parts (DC and AC cases) to make your reasoning easy to follow.

Question 510. A capacitor made of two parallel plates each of plate area A and separation d, is being charged by an external a.c. source. Show that the displacement current inside the capacitor is same as the current charging the capacitor.
Answer: Let the alternating voltage applied across the parallel plate capacitor of capacitance \( C = \frac{\varepsilon_0 A}{d} \) be: \[ V = V_0 \sin\omega t \]
**1. Calculating the Conduction Current (\( I_C \)):**
The instantaneous conduction current charging the plates is: \[ I_C = \frac{dq}{dt} = \frac{d}{dt} (C V) \] Substituting \( V \): \[ I_C = C \frac{d}{dt} (V_0 \sin\omega t) = C V_0 \omega \cos\omega t \] Using \( C = \frac{\varepsilon_0 A}{d} \): \[ I_C = \frac{\varepsilon_0 A}{d} V_0 \omega \cos\omega t \]
**2. Calculating the Displacement Current (\( I_D \)):**
The electric field \( E \) between the plates is \( E = \frac{V}{d} \). The electric flux is: \[ \Phi_E = E A = \frac{V A}{d} \] Using the definition of displacement current: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d}{dt} \left( \frac{V A}{d} \right) \] Substituting \( V = V_0 \sin\omega t \): \[ I_D = \frac{\varepsilon_0 A}{d} \frac{d}{dt} (V_0 \sin\omega t) = \frac{\varepsilon_0 A}{d} V_0 \omega \cos\omega t \]
Comparing both equations, we find: \[ I_C = I_D \] This proves that the displacement current inside the capacitor is exactly equal to the conduction current charging it.
In simple words: By calculating the real current flowing into the plates using the voltage equation, and then calculating the displacement current from the changing electric field, we get the exact same mathematical formula. This proves they are identical.

Exam Tip: Show both step-by-step derivatives of \( \sin\omega t \) clearly to make your mathematical proof easy to grade.

Question 511. Write the expression for the generalized Ampere’s circuital law. Through a suitable example, explain the significance of time dependent term. 
Answer: The generalized Ampere's Circuital Law (also called the Ampere-Maxwell Law) is expressed as: \[ \oint \vec{B} \cdot \vec{dl} = \mu_0 \left( I_C + I_D \right) = \mu_0 \left( I_C + \varepsilon_0 \frac{d\Phi_E}{dt} \right) \] where \( I_C \) is the conduction current and \( I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \) is the time-dependent displacement current term.

**Significance of the Time-Dependent Term (Example of a Charging Capacitor):**
Consider a capacitor being charged by a battery. If we draw a flat circular loop around the wire leading to the capacitor, Ampere's law gives a magnetic field because it encloses a conduction current \( I_C \). However, if we stretch this loop into a pot-like shape passing *between* the capacitor plates, it encloses no conduction current (\( I_C = 0 \)), which would suggest the magnetic field is zero. This contradiction is resolved by the time-dependent term \( \varepsilon_0 \frac{d\Phi_E}{dt} \). Inside the gap, the changing electric flux creates a displacement current that matches \( I_C \), ensuring the magnetic field remains continuous and consistent across both regions.
In simple words: The modified law adds a term for changing electric fields. This term is crucial because, inside a charging capacitor's gap, it replaces the missing wire current and ensures that magnetic fields don't mathematically disappear where the wires end.

Exam Tip: Clearly sketch or describe the "pot-like surface" between capacitor plates to illustrate this classic contradiction and its resolution.

Question 512. What are electromagnetic waves ? Are these waves transverse or longitudinal ?
Answer: Electromagnetic waves are self-propagating waves produced by accelerating or oscillating charged particles. They consist of sinusoidally oscillating electric (\( \vec{E} \)) and magnetic (\( \vec{B} \)) field vectors that oscillate perpendicular to each other and also perpendicular to the direction of wave propagation.
These waves are strictly **transverse** in nature.
In simple words: Electromagnetic waves are travelling ripples of electric and magnetic fields created by moving charges. They are transverse waves, meaning their fields vibrate sideways to the direction the wave is travelling.

Exam Tip: Define EM waves by mentioning that both fields oscillate perpendicular to each other and to the wave's path.

Question 513. (i) How are electromagnetic waves produced ? Explain.
(ii) What is the source of energy of these waves ?
Answer: The explanations are:
(i) **Production of EM Waves:**
Electromagnetic waves are generated by accelerating or oscillating electric charges. An oscillating charge creates a time-varying electric field in its vicinity. This changing electric field induces a time-varying magnetic field perpendicular to it. In turn, this changing magnetic field induces a changing electric field, and this continuous regeneration propagates through space as an electromagnetic wave.

(ii) **Source of Energy:**
The source of energy for electromagnetic waves is the mechanical or electrical energy of the accelerating or oscillating charge that produced them.
In simple words: (i) When a charge wiggles, it creates a changing electric field, which creates a changing magnetic field, which then creates another electric field. This endless chain reaction travels through space as a wave. (ii) The energy of the wave comes directly from the kinetic energy of the wiggling charge.

Exam Tip: Use the cycle of "changing electric field \( \rightarrow \) changing magnetic field \( \rightarrow \) changing electric field" to explain wave propagation clearly.

Question 514. What oscillates in electromagnetic waves ? 
Answer: In an electromagnetic wave, the electric field vector (\( \vec{E} \)) and the magnetic field vector (\( \vec{B} \)) oscillate sinusoidally in space and time.
In simple words: It is the strength of the electric and magnetic fields that vibrates up and down in an EM wave.

Exam Tip: Specify that both the electric and magnetic field vectors are what oscillate.

Question 515. What is the phase relationship between oscillating electric and magnetic fields in an em wave ? 
Answer: The oscillating electric field (\( \vec{E} \)) and magnetic field (\( \vec{B} \)) vectors in an electromagnetic wave are in the **same phase**. This means they reach their maximum values and their minimum (zero) values at the exact same moments in time and coordinates in space.
In simple words: They are perfectly in sync. When the electric field peaks, the magnetic field peaks too; when one drops to zero, the other does as well.

Exam Tip: Use the phrase "in the same phase" to describe this synchronization.

Question 516. What is the frequency of em waves produced by oscillating charge of frequency \nu ? 
Answer: The frequency of the electromagnetic waves produced is exactly equal to the frequency of the oscillating charge that generates them: \[ f_{\text{wave}} = \nu \]
In simple words: The wave vibrates at the exact same rate as the charge that is wiggling to produce it.

Exam Tip: State clearly that the wave frequency is identical to the source charge oscillation frequency.

Question 517. When can a charge acts as a source of em wave ?
Answer: An electric charge can act as a source of electromagnetic waves only when it undergoes **acceleration** (such as linear acceleration, circular motion, or continuous harmonic oscillation). A charge moving at a constant, uniform velocity cannot produce EM waves.
In simple words: A charge only radiates EM waves when it is accelerating, slowing down, turning, or vibrating. If it moves in a straight line at a steady speed, it won't produce any waves.

Exam Tip: Emphasize that "accelerating or oscillating" is the mandatory condition for a charge to emit radiation.

Question 518. Write the relation for the speed of electromagnetic waves in terms of the amplitudes of electric and magnetic fields.
Answer: The speed of electromagnetic waves \( c \) in free space is defined by the ratio of the amplitude of the electric field vector \( E_0 \) to the amplitude of the magnetic field vector \( B_0 \): \[ c = \frac{E_0}{B_0} \]
In simple words: The speed of light is simply the peak strength of the electric field divided by the peak strength of the magnetic field.

Exam Tip: Write the ratio clearly as \( c = \frac{E_0}{B_0} \) and specify that \( E_0 \) and \( B_0 \) are peak amplitudes.

Question 519. Write the expression for speed of electromagnetic waves in a medium of electrical permittivity \varepsilon and magnetic permeability \mu.
Answer: The speed \( v \) of electromagnetic waves inside a material medium with electrical permittivity \( \varepsilon \) and magnetic permeability \( \mu \) is given by: \[ v = \frac{1}{\sqrt{\mu \varepsilon}} \] In free space (vacuum), this is written as: \[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \]
In simple words: The speed of the wave in any material depends on how easily the material polarizes electrically and magnetically, calculated using one over the square root of their product.

Exam Tip: Show both the general medium formula \( v = \frac{1}{\sqrt{\mu\varepsilon}} \) and the free-space limit \( c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \) to show comprehensive knowledge.

Question 520. What is meant by the transverse nature of electromagnetic waves ? 
Answer: The transverse nature means that the electric field vector (\( \vec{E} \)) and the magnetic field vector (\( \vec{B} \)) oscillate perpendicular to each other, and both oscillate perpendicular to the direction in which the wave is traveling (propagating).
In simple words: It means the field vibrations occur at right angles to the direction the wave is moving, like a wave travelling down a plucked string.

Exam Tip: State that \( \vec{E} \perp \vec{B} \perp \vec{v} \) where \( \vec{v} \) is the velocity of the wave.

Question 530. How are the directions of the electric and magnetic field vectors in an em wave are related to each other and to the direction of propagation of the em waves ? 
Answer: The electric field vector \( \vec{E} \) and magnetic field vector \( \vec{B} \) are perpendicular to each other, and their vector cross product \( \vec{E} \times \vec{B} \) points in the direction of wave propagation.
In simple words: The electric and magnetic fields are at 90 degrees to each other, and their combined cross product points straight along the path the wave is travelling.

Exam Tip: Mention that the direction of propagation is given by the unit vector along \( \vec{E} \times \vec{B} \).

Question 531. In which directions do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the x-axis ? 
Answer: When the electromagnetic wave propagates along the x-axis:
* The electric field vector \( \vec{E} \) oscillates along the y-axis, and the magnetic field vector \( \vec{B} \) oscillates along the z-axis.
* Alternatively, \( \vec{E} \) can oscillate along the z-axis, while \( \vec{B} \) oscillates along the y-axis.
In simple words: If the wave travels along the X-axis, the electric and magnetic fields must oscillate along the Y and Z axes (or vice versa) to remain perpendicular.

Exam Tip: Give both possible perpendicular combinations (E along y / B along z, or E along z / B along y) to write a complete answer.

Question 532. Write mathematical expression for electric and magnetic fields of an electromagnetic wave propagating along z-axis.
Answer: For a plane electromagnetic wave propagating along the positive z-axis, the oscillating fields can be mathematically written as: \[ E_x = E_0 \sin(K z - \omega t) \hat{i} \] \[ B_y = B_0 \sin(K z - \omega t) \hat{j} \] where \( E_0 \) and \( B_0 \) are field amplitudes, \( K = \frac{2\pi}{\lambda} \) is the wave vector, and \( \omega = 2\pi f \) is the angular frequency.
In simple words: We can write their oscillations as sine waves. If the wave travels along Z, the electric field oscillates along X and the magnetic field oscillates along Y.

Exam Tip: Make sure the propagation term inside the sine function is written as \( (Kz - \omega t) \) to show propagation along the z-axis.

Question 533. Draw a sketch of linearly polarized em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields. 
Answer: Below is the diagram showing the transverse, linearly polarized electromagnetic wave propagating along the positive z-axis. The electric field oscillates in the y-z plane (along the y-axis) and the magnetic field oscillates in the x-z plane (along the x-axis).

In simple words: This sketch shows the electric field wave undulating vertically along the Y-axis and the magnetic field wave undulating horizontally along the X-axis, with both traveling forward along the Z-axis.

Exam Tip: Clearly label all three axes (X, Y, Z) and the respective fields (\( \vec{E} \), \( \vec{B} \)) to ensure full marks.

Question 534. Write the expression for the energy density of an electromagnetic wave propagating in free space.
Answer: The total average energy density \( u \) of an electromagnetic wave traveling in free space is the sum of the average energy density of the electric field (\( u_E \)) and the magnetic field (\( u_B \)): \[ u = u_E + u_B = \frac{1}{2} \varepsilon_0 E^2 + \frac{B^2}{2 \mu_0} \] Since both energy densities are equal (\( u_E = u_B \)), the expression can also be written in terms of only one of the fields: \[ u = \varepsilon_0 E_{\text{rms}}^2 = \frac{B_{\text{rms}}^2}{\mu_0} \]
In simple words: The energy stored in the wave is shared equally between its electric and magnetic parts. The total energy density is the sum of these two parts.

Exam Tip: Show that \( u_E = u_B \) to demonstrate a complete understanding of how energy is divided in EM waves.

Question 535. State any four properties of electromagnetic waves. 
Answer: Four key properties of electromagnetic waves are:
1. **No Medium Required:** They do not require any material medium for their propagation and can travel through a vacuum.
2. **Transverse Nature:** The electric and magnetic field oscillations are perpendicular to each other and to the wave's direction of propagation.
3. **No Deflection:** Being uncharged, they are completely unaffected by external electric or magnetic fields.
4. **Constant Speed:** In a vacuum, all electromagnetic waves travel at the same speed, which is the speed of light (\( c \approx 3 \times 10^8\text{ m/s} \)).
In simple words: They don't need air or material to travel, they are transverse, they aren't bent by magnets or electric plates, and they all travel at the speed of light in a vacuum.

Exam Tip: Present these properties as a numbered list to make it clean and easy for the examiner to award full marks.

Question 536. Do the electromagnetic waves carry energy and momentum ?
Answer: Yes, electromagnetic waves carry both energy and momentum as they propagate through space. This is demonstrated when they strike a surface and exert radiation pressure on it.
In simple words: Yes, they do. Just like physical objects, traveling light waves carry real energy and momentum.

Exam Tip: Answer with a direct "Yes" before explaining the physical proof (radiation pressure).

Question 537. How can we show that em waves carry momentum ?
Answer: When an electromagnetic wave strikes a surface containing electric charges (like a metal sheet), the oscillating electric and magnetic fields of the wave exert forces on these charges, setting them into motion. As the charges absorb energy and gain momentum from the wave, they prove that the wave itself carries momentum.
If the wave transfers a total energy \( U \) to a completely absorbing surface, the momentum \( p \) delivered is: \[ p = \frac{U}{c} \]
In simple words: When light hits a surface, its oscillating fields push on the surface's electrons, transferring kinetic energy and momentum to them. This physical push is the direct proof that the waves carry momentum.

Exam Tip: Write down the momentum-energy transfer formula \( p = \frac{U}{c} \) to make your answer highly rigorous.

Question 538. Why is the amount of the momentum transferred by the EM waves incident on the surface so small ? 
Answer: The momentum \( p \) transferred by an electromagnetic wave carrying energy \( U \) is given by: \[ p = \frac{U}{c} \] Since the speed of light \( c \) in the denominator is exceptionally large (\( 3 \times 10^8\text{ m/s} \)), the resulting value of momentum \( p \) is extremely small for any typical, everyday wave energy.
In simple words: Because light travels so fast, its speed divides the wave's energy in the momentum formula, making the actual physical push (momentum) incredibly tiny.

Exam Tip: Clearly state that the extremely large value of \( c \) in the denominator of \( p = U/c \) is the physical cause.

Question 539. An em wave exerts pressure on the surface on which it is incident. Justify. 
Answer: Since electromagnetic waves carry momentum \( p \), they deliver this momentum to any surface they hit. According to Newton's second law, this transfer of momentum over time creates a force: \[ F = \frac{dp}{dt} \] The force exerted per unit area of the surface is called radiation pressure: \[ P = \frac{F}{A} \] This force per unit area is the physical pressure exerted by the wave.
In simple words: Because light has momentum, hitting a surface is like a stream of tiny particles bouncing off it. This continuous transfer of momentum exerts a real force over the area, creating radiation pressure.

Exam Tip: Use Newton's second law (\( F = dp/dt \)) to link wave momentum to the concept of pressure.

Question 540. Figure shows a capacitor made of two circular plates. The capacitor is being charged by an external source. The charging current is constant and equal to 0.15 A.
(a) What is the displacement current across the plates.
(b) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer: The calculations and explanations are:
(a) The displacement current \( I_D \) between the plates of a capacitor is always equal to the conduction current \( I_C \) flowing through the connecting wires during the charging process. Therefore, the displacement current is: \[ I_D = 0.15\text{ A} \]
(b) Yes, Kirchhoff's first rule (junction rule) remains valid at each plate of the capacitor.
**Explanation:**
When we include both conduction current and displacement current, the total current is continuous. The conduction current \( I_C \) arriving at a plate is exactly equal to the displacement current \( I_D \) leaving the plate into the gap, meaning the net current entering or leaving any junction plate is zero.

In simple words: (a) The displacement current is exactly equal to the charging wire current, which is 0.15 Amperes. (b) Yes, the junction rule still works because the incoming wire current matches the outgoing virtual current in the gap, keeping the total flow continuous.

Exam Tip: Emphasize that the continuity of total current (\( I_C = I_D \)) is what saves Kirchhoff's junction rule at the capacitor plates.

Question 541. Which physical quantity, if any, has the same value for the waves belonging to the different parts of the electromagnetic spectrum ?
Answer: The physical quantity that remains constant for all components of the electromagnetic spectrum in a vacuum is **velocity** (equal to \( c = 3 \times 10^8\text{ m/s} \)).
In simple words: All different types of electromagnetic waves, from radio waves to gamma rays, travel at the exact same speed in a vacuum.

Exam Tip: Specify that this constancy of velocity holds true specifically "in a vacuum."

Question 542. Name the physical quantity which remains same for microwaves of wavelength 1mm and UV radiations of 1600 A0 in vacuum. 
Answer: The physical quantity that remains identical is the **velocity** (speed of propagation), which is: \[ c = 3 \times 10^8\text{ m/s} \]
In simple words: Both microwaves and ultraviolet rays travel at the speed of light (\( 3 \times 10^8 \) meters per second) when moving through a vacuum.

Exam Tip: Write down the numerical value of the speed of light along with the name of the physical quantity.

Question 543. What is the ratio of speed of infrared and ultraviolet rays in vacuum ?
Answer: Since all electromagnetic waves travel with the same speed \( c \) in a vacuum, the ratio of the speed of infrared rays to that of ultraviolet rays is: \[ 1 : 1 \]
In simple words: Because both waves travel at the exact same speed in empty space, the ratio of their speeds is simply 1 to 1.

Exam Tip: State the physical reason (same velocity \( c \) in a vacuum) before writing the final ratio.

Question 544. Give the ratio of velocities of wavelengths 4000 A0 and 8000 A0 in vacuum ? 
Answer: In a vacuum, the velocity of any electromagnetic wave is independent of its wavelength. Therefore, the ratio of their velocities is: \[ 1 : 1 \]
In simple words: The speed of an electromagnetic wave in a vacuum doesn't depend on its wavelength, so both waves travel at the same speed, giving a 1 to 1 ratio.

Exam Tip: Reiterate that vacuum speed is constant for all electromagnetic wavelengths.

Question 545. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations & write the range of their frequency. 
Answer: The details are:
* **Name of Radiations:** **Ultraviolet (UV) radiations** (which are emitted by the high-temperature welding arcs).
* **Frequency Range:** Approximately \( 10^{14}\text{ Hz} \) to \( 10^{16}\text{ Hz} \) (or more specifically, \( 8 \times 10^{14}\text{ Hz} \) to \( 3 \times 10^{16}\text{ Hz} \)).
In simple words: Welding arcs produce harmful ultraviolet (UV) rays. Welders wear special masks to block these rays, which vibrate at a high frequency of around \( 10^{14} \) to \( 10^{16} \) Hertz.

Exam Tip: Be sure to mention both the name of the radiation (Ultraviolet) and its standard textbook frequency range.

Question 546. Why are microwaves found useful for the radar systems in aircraft navigation ?
OR
State the reason why microwaves are best suited for long distance transmission of signals ? 
Answer: Due to their relatively short wavelengths, microwaves can travel in narrow, highly concentrated beams without showing significant diffraction (bending) around obstacles. This allows them to travel long distances in straight lines through the atmosphere with minimal signal loss, making them ideal for radar detection and satellite communications.
In simple words: Microwaves have very short wavelengths, meaning they travel in straight, laser-like beams without spreading out or bending around obstacles. This keeps the signal strong and clear over long distances.

Exam Tip: Use the key phrase "short wavelength prevents diffraction" to explain why the beams remain narrow and directional.

Question 547. Why is the thin ozone layer on the top of stratosphere is crucial for human survival ? Identify to which part of electromagnetic spectrum does this radiation belong and write one important application of the radiation.
Answer: The ozone layer absorbs highly energetic, harmful solar ultraviolet (UV) radiation, preventing it from reaching the Earth's surface where it could cause skin cancers, cataracts, and ecological damage.
* **Identification:** These waves belong to the **Ultraviolet (UV) region** of the electromagnetic spectrum.
* **Application:** They are widely used in water purification systems to destroy bacteria, and in forensic laboratories to detect forged documents.
In simple words: The ozone layer acts as a shield, blocking dangerous ultraviolet (UV) rays from the sun that can cause skin cancer. These same UV rays are used in water purifiers to kill germs.

Exam Tip: Break your answer into clear, labeled parts: Importance, Identification, and Application to help the examiner grade it easily.

Question 548. How are infrared rays produced ? Why are these referred to as “ heat waves? Write their three important uses. Name the radiations which are next to these radiations in the electromagnetic spectrum having (a) shorter wavelength (b) longer wavelength.
Answer: The details are:
* **Production:** Infrared waves are generated by the thermal vibrations of atoms and molecules in hot bodies.
* **Why called "heat waves":** They are absorbed readily by water molecules in most objects, increasing their thermal kinetic energy and causing the temperature of the object to rise.
* **Three Important Uses:**
1. Used in night-vision cameras and photography through dense fog.
2. Used in physical therapy to treat muscular strains and joint pain.
3. Used in household remote controls for electronic appliances.
* **Adjacent Radiations:**
(a) Radiations with a shorter wavelength: **Visible light**.
(b) Radiations with a longer wavelength: **Microwaves**.
In simple words: Infrared rays are made by jiggling atoms in hot objects. They are called heat waves because water absorbs them easily, making things warm up. We use them in TV remotes and night-vision goggles. Visible light has shorter waves, and microwaves have longer waves.

Exam Tip: Be careful to list the adjacent bands correctly: visible light lies on the shorter wavelength side, and microwaves on the longer side.

Question 549. What role does infra radiation play in (i) maintain the Earth’s warmth, and (ii) Physical therapy ?
Answer: The roles are defined below:
**(i) Maintaining Earth's Warmth:** The Earth's surface absorbs solar radiation and re-radiates it as longer-wavelength infrared waves. Greenhouse gases (like \( \text{CO}_2 \) and water vapor) trap these outgoing infrared waves, keeping the heat from escaping into space and maintaining a habitable planetary temperature.
**(ii) Physical Therapy:** Infrared radiation is easily absorbed by the water molecules in human tissue. Upon absorption, the thermal motion of the molecules increases, heating the affected muscle tissue. This localized heating relaxes muscles and relieves strain.
In simple words: (i) The Earth warms up and radiates heat back as infrared rays, which get trapped by greenhouse gases to keep our planet warm. (ii) Infrared rays penetrate skin and warm up muscles, which relaxes them and treats pain.

Exam Tip: Mention "greenhouse effect" for the first part and "absorption by water molecules" for the second part.

Question 550. If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now ? Explain. 
Answer: The average surface temperature would be **significantly lower** than it is currently.
**Explanation:** Without an atmosphere, there would be no greenhouse gases to trap the infrared heat radiated by the Earth's surface. As a result, all the heat would escape into outer space, leaving the planet freezing.
In simple words: It would be much colder because there would be no greenhouse gases to act as a blanket and trap the heat, causing all warmth to escape into space.

Exam Tip: Use the term "absence of greenhouse effect" as the central explanation for the drop in temperature.

Question 551. State clearly how a microwave oven works to heat up a food item containing water molecules?
Answer: In a microwave oven, the frequency of the generated microwaves is selected to match the natural resonant frequency of water molecules (approximately \( 3\text{ GHz} \)). This resonance allows efficient energy transfer from the waves to the rotational kinetic energy of the water molecules in the food, raising its temperature rapidly.
In simple words: Microwaves vibrate at the exact frequency water molecules naturally spin. This causes the water inside the food to vibrate violently, generating heat from the inside out.

Exam Tip: Use the word "resonance" to explain how energy is efficiently transferred from the waves to the water molecules.

Question 552. Which segment of electromagnetic waves has highest frequency ? How are these waves produced ? Give one use of these waves.
Answer: The details are:
* **Highest Frequency Segment:** **Gamma (\( \gamma \)) rays**.
* **Production:** They are produced during the radioactive decay of unstable atomic nuclei or in high-energy nuclear reactions.
* **Use:** They are widely used in medicine for cancer radiation therapy to destroy malignant tumor cells.
In simple words: Gamma rays have the highest frequency. They are created by radioactive nuclei during nuclear decay and are used in hospitals to target and destroy cancer cells.

Exam Tip: Name "nuclear decay" or "radioactive decay" as the production source to secure full marks.

Question 553. Which em waves lie near the high frequency end of visible part of em spectrum ? Give its one use. In what way This component of light has harmful effects on humans ?
Answer: The waves lying near the high-frequency limit of the visible spectrum are **Ultraviolet (UV) rays**.
* **One Use:** They are utilized in LASIK eye surgeries and in specialized germicidal lamps to sterilize water.
* **Harmful Effects:** Exposure to direct UV radiation can cause painful sunburns, trigger skin cancers, and damage the retinas of human eyes.
In simple words: Ultraviolet (UV) rays sit right next to the high-frequency end of visible light. They are used in laser eye surgery, but overexposure can cause sunburns, skin cancer, and eye damage.

Exam Tip: Clearly state that UV rays sit just beyond the violet end of visible light.

Question 554. Which of the following electromagnetic radiations has least frequency : 
UV radiations, X-rays, Microwaves
Answer: Among the choices, **Microwaves** have the lowest (least) frequency.
In simple words: Out of the three, microwaves vibrate the slowest, meaning they have the lowest frequency.

Exam Tip: Remember that order of frequency from lowest to highest is Microwaves < UV < X-rays.

Question 555. Which of the following has the shortest wavelength : 
Microwaves, Ultraviolet rays, X-rays
Answer: Among the options provided, **X-rays** possess the shortest wavelength.
In simple words: X-rays have the shortest wavelength among these three because they have the highest frequency and energy.

Exam Tip: Note that shortest wavelength corresponds directly to the highest frequency.

Question 556. Arrange the following electromagnetic waves in order of increasing frequency :
\gamma-rays, microwaves, infrared rays and Ultraviolet rays
Answer: In order of increasing frequency (lowest frequency first): \[ \text{Microwaves} < \text{Infrared rays} < \text{Ultraviolet rays} < \gamma\text{-rays} \]
In simple words: From lowest to highest frequency, the order is: Microwaves, then Infrared, then Ultraviolet, and finally Gamma rays.

Exam Tip: Use inequality symbols (\( < \)) to display your sorted list clearly.

Question 557. Arrange the following electromagnetic waves in order of decreasing frequency :
x-rays, \gamma-rays, microwaves, UV rays and infrared rays
Answer: In order of decreasing frequency (highest frequency first): \[ \gamma\text{-rays} > \text{X-rays} > \text{UV rays} > \text{Infrared rays} > \text{Microwaves} \]
In simple words: From highest to lowest frequency, the order is: Gamma rays, X-rays, Ultraviolet, Infrared, and Microwaves.

Exam Tip: Since the question asks for "decreasing order," use the greater-than symbol (\( > \)) or list them with a clear descending label.

Question 558. Arrange the following em waves in order of their increasing wavelength : 
\gamma-rays, Microwaves, X-rays, U.V. rays and Radio waves
Answer: In order of increasing wavelength (shortest wavelength first): \[ \gamma\text{-rays} < \text{X-rays} < \text{U.V. rays} < \text{Microwaves} < \text{Radio waves} \]
In simple words: From shortest to longest wavelength, the order is: Gamma rays, X-rays, Ultraviolet, Microwaves, and Radio waves.

Exam Tip: Remember that wavelength order is the exact opposite of frequency order.

Question 559. Arrange the following electromagnetic waves in decreasing order of wavelength : 
\gamma-rays, infrared rays, x-rays and microwaves
Answer: In order of decreasing wavelength (longest wavelength first): \[ \text{Microwaves} > \text{Infrared rays} > \text{X-rays} > \gamma\text{-rays} \]
In simple words: From longest to shortest wavelength, the order is: Microwaves, Infrared, X-rays, and Gamma rays.

Exam Tip: Double check whether the question asks for "increasing" or "decreasing" order before finalizing your sequence.

Question 560. Name the following constituent radiations of electromagnetic spectrum which-
(i) are used in satellite communication/in radar and geostationary satellite 
(ii) are used for studying crystal structure of solids 
(iii) are similar to the radiations emitted during decay of radioactive nuclei
(iv) used for water purification/ are absorbed from sunlight by ozone layer
Answer: The identifications are:
(i) **Microwaves** (due to their directional, narrow-beam propagation characteristics).
(ii) **X-rays** (due to their short wavelengths being comparable to interatomic spacing).
(iii) **Gamma (\( \gamma \)) rays** (which are physically identical to nuclear decay emissions).
(iv) **Ultraviolet (UV) rays** (which are absorbed by ozone and used to sterilize water).
In simple words: (i) Microwaves are used for satellites. (ii) X-rays are used to study crystals. (iii) Gamma rays are identical to radioactive emissions. (iv) UV rays purify water and are absorbed by ozone.

Exam Tip: Match each sub-part with its exact number (i to iv) to present a clean, easily gradeable solution.

Question 561. Name the following constituent radiations of electromagnetic spectrum which- 
(i) has its wavelength range between 390 nm to 770 nm 
(ii) produce intense heating effect/ used in warfare to look through fog
(iii) are used for radar systems used in aircraft navigation 
Answer: The identifications are:
(i) **Visible Light** (the band detected by human eyes).
(ii) **Infrared rays** (referred to as heat waves, capable of penetrating fog).
(iii) **Microwaves** (used for highly directional radar beams).
In simple words: (i) Visible light is what we see. (ii) Infrared rays create heat and let us see through fog. (iii) Microwaves are used for aircraft radar.

Exam Tip: Be precise with wavelength limits; \( 390\text{ nm} \) to \( 770\text{ nm} \) is the exact spectrum of visible light.

Question 562. Name the following constituent radiations of electromagnetic spectrum which-
(i) are adjacent to the low frequency end of electromagnetic spectrum 
(ii) produced by nuclear reactions/used to destroy cancer cells/treatment of cancer 
(iii) produced by bombarding a metal target by high speed electrons. 
(iv) maintains the earth’s warmth/ used in remote sensing
Answer: The identifications are:
(i) **Microwaves** (which sit adjacent to the low-frequency radio waves).
(ii) **Gamma (\( \gamma \)) rays** (highly energetic waves used in cancer therapies).
(iii) **X-rays** (produced by electron deceleration on target metals).
(iv) **Infrared rays** (responsible for the greenhouse effect and thermal imaging).
In simple words: (i) Microwaves are next to radio waves. (ii) Gamma rays treat cancer. (iii) X-rays are made by shooting electrons at metal targets. (iv) Infrared rays keep the Earth warm.

Exam Tip: Make sure you name the correct specific radiation for each production mechanism described.

Question 563. Which constituent radiations of electromagnetic spectrum is used - 
(i) in Radar
(ii) in photographs of internal parts of human body/as a diagnostic tool in medicine 
(iii) for taking photographs of sky, during night and fog conditions.
(iv) has the largest penetrating power 
Give reason for your answer in each case.
Answer: The details are:
(i) **Microwaves:** Because they have short wavelengths, allowing them to travel in straight lines through the atmosphere with minimal scattering or diffraction.
(ii) **X-rays:** Because they can easily penetrate soft tissue (flesh) but are absorbed by denser materials like bones, leaving a clear shadow image.
(iii) **Infrared rays:** Because they have longer wavelengths that easily pass through dense fog particles without getting scattered as much as visible light.
(iv) **Gamma (\( \gamma \)) rays:** Because they possess the highest frequency and shortest wavelength, giving them the highest energy per photon and thus the largest penetrating power.
In simple words: (i) Microwaves are used in radar because they travel in straight lines. (ii) X-rays show bones because they pass through flesh but are blocked by bone. (iii) Infrared rays can cut through fog because they scatter less. (iv) Gamma rays have the highest energy, allowing them to penetrate deep into materials.

Exam Tip: State both the name of the radiation and the physical reasoning behind its application to get full marks.

Question 564. Electromagnetic waves with wavelengths-
(i) \lambda_1 are used to treat muscular strain 
(ii) \lambda_2 are used by a F.M. radio station for broadcasting
(iii) \lambda_3 are used to detect fractures in bones 
(iv) \lambda_4 are absorbed by ozone layer of the atmosphere
Identify the name and part of electromagnetic spectrum to which these radiations belong. Arrange these wavelengths in order of magnitude.
Answer: The details are:
* **Identification:**
(i) \( \lambda_1 \): **Infrared rays** (used for deep heat treatment).
(ii) \( \lambda_2 \): **Radio waves** (used for FM signal broadcasting).
(iii) \( \lambda_3 \): **X-rays** (used to see bone structures).
(iv) \( \lambda_4 \): **Ultraviolet (UV) rays** (absorbed by stratospheric ozone).

* **Order of Magnitude (Wavelengths):**
Arranging from longest to shortest: \[ \lambda_2 > \lambda_1 > \lambda_4 > \lambda_3 \] (Radio waves \( > \) Infrared \( > \) Ultraviolet \( > \) X-rays).
In simple words: (i) Infrared rays treat muscles. (ii) Radio waves broadcast FM. (iii) X-rays detect bone fractures. (iv) UV rays are absorbed by ozone. Sorted from longest to shortest wavelength, the order is Radio, Infrared, UV, and X-rays.

Exam Tip: Ensure you explicitly write down the sorted wavelength relation \( \lambda_2 > \lambda_1 > \lambda_4 > \lambda_3 \) at the end of your answer.

Question 565. Identify the electromagnetic waves whose wavelength vary as and also write one use for each. 
(i) 10^{-12}\text{ m} < \lambda < 10^{-8}\text{ m} \quad (ii) 10^{-3}\text{ m} < \lambda < 10^{-1}\text{ m}
Answer: The identifications are:
(i) **X-rays:** They are widely used in medicine to detect fractures in bones and internal structural issues.
(ii) **Microwaves:** They are used in radar systems for aircraft navigation and speed detection.
In simple words: (i) The first range is X-rays, which are used to scan bones. (ii) The second range is microwaves, used in radar and cooking.

Exam Tip: Identify the correct category first before writing a standard, well-known application.

Question 566. Identify the electromagnetic waves whose wavelength vary as and also write one use for each. 
(i) 10^{-11}\text{ m} < \lambda < 10^{-14}\text{ m} \quad (ii) 10^{-4}\text{ m} < \lambda < 10^{-6}\text{ m}
Answer: The identifications are:
(i) **Gamma (\( \gamma \)) rays:** Used in oncology for radiation therapy to target and destroy cancer cells.
(ii) **Infrared rays:** Used in thermal cameras and remote controls for home electronic devices.
In simple words: (i) The first range represents gamma rays, used to treat cancer. (ii) The second range represents infrared rays, used in remote controls and heat sensors.

Exam Tip: Pay attention to negative exponents: smaller negative exponents like \( 10^{-6} \) mean longer wavelengths compared to \( 10^{-14} \).

Question 567. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is ID = \varepsilon_0 \frac{d\Phi_E}{dt} 
where \Phi_E is the electric flux produced during charging of the capacitor plates.
Answer: Let \( A \) be the area of each plate of the capacitor, and \( q \) be the charge on the plates at any instant during the charging process.
The electric field \( E \) between the plates is given by: \[ E = \frac{\sigma}{\varepsilon_0} = \frac{q}{\varepsilon_0 A} \] The electric flux \( \Phi_E \) passing through the space between the plates is: \[ \Phi_E = E A = \left(\frac{q}{\varepsilon_0 A}\right) A = \frac{q}{\varepsilon_0} \] Multiplying both sides by \( \varepsilon_0 \): \[ q = \varepsilon_0 \Phi_E \] Taking the derivative with respect to time \( t \) on both sides: \[ \frac{dq}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt} \] Since the current within the plates is defined as the displacement current \( I_D = \frac{dq}{dt} \): \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} \] This mathematically proves the expression.
In simple words: The electric field between the plates is directly linked to the charge. By calculating the electric flux and taking its rate of change over time, we show that the virtual current (displacement current) inside is indeed \( \varepsilon_0 \) times the change in flux.

Exam Tip: Show the substitution of \( E = \frac{q}{\varepsilon_0 A} \) clearly to construct a logical, step-by-step proof.

Question 568. Show that in the process of charging a capacitor, displacement current is always equal to conduction current. 
Answer: During the charging process, let \( q \) be the instantaneous charge accumulating on the capacitor plates.
The conduction current \( I \) flowing through the connecting wires is: \[ I = \frac{dq}{dt} \] The electric field \( E \) between the circular plates of area \( A \) is \( E = \frac{q}{\varepsilon_0 A} \), giving the electric flux as: \[ \Phi_E = E A = \frac{q}{\varepsilon_0} \] Using the formula for displacement current \( I_D \): \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d}{dt} \left( \frac{q}{\varepsilon_0} \right) = \frac{dq}{dt} \] Comparing both equations: \[ I_D = I \] This shows that the displacement current within the gap is always identical to the conduction current in the wires.
In simple words: By taking the time derivative of the electric flux inside the capacitor, we get the exact same formula (\( dq/dt \)) as the wire current. This proves they are mathematically equal.

Exam Tip: Start by writing down the definitions of both currents to show how they naturally equate.

Question 569. Why does a galvanometer when connected in series with a capacitor show a momentary deflection, when it is Being charged or discharged ? How does this information lead to modify the Ampere’s circuital law ? Hence write the generalized expression of Ampere’s circuital law. 
Answer: The explanations and expressions are:
* **Why Momentary Deflection occurs:**
During the active charging or discharging of a capacitor, the electric field in the gap varies with time. This changing field sets up a displacement current \( I_D \) across the plates, completing the circuit loop and causing a brief conduction current to flow through the wires and deflect the galvanometer. Once fully charged or discharged, the field stops changing, \( I_D \) drops to zero, and the deflection vanishes.
* **How it modifies Ampere's Law:**
Classic Ampere's Law states \( \oint \vec{B} \cdot \vec{dl} = \mu_0 I_C \). If we apply this to a flat loop around the wire, we get a non-zero magnetic field. But if we stretch the loop into a pot-like shape passing *between* the plates, the enclosed conduction current is zero, leading to a contradiction. To resolve this, Maxwell introduced the displacement current term, making the law continuous across all surfaces.
* **Generalized Ampere's Law:** \[ \oint \vec{B} \cdot \vec{dl} = \mu_0 \left( I_C + I_D \right) = \mu_0 \left( I_C + \varepsilon_0 \frac{d\Phi_E}{dt} \right) \]
In simple words: The needle only moves while the capacitor is active because the changing electric field creates a temporary virtual current. This gap in classical physics forced scientists to add a "displacement current" term to Ampere's law so that magnetic field calculations remain continuous across any surface shape.

Exam Tip: State both the contradiction (fields disappearing between plates) and its resolution to provide a complete answer.

Question 570. Write the generalized expression for Ampere’s circuital law in terms of the conduction current and displacement current. Mention the situation when there is : 
(i) only conduction current and no displacement current
(ii) only displacement current and no conduction current
Answer: The generalized Ampere's circuital law is: \[ \oint \vec{B} \cdot \vec{dl} = \mu_0 (I_C + I_D) \]
(i) **Only Conduction Current (\( I_D = 0 \)):**
This occurs in a straight, steady current-carrying wire where the electric field is constant over time, making \( \frac{d\Phi_E}{dt} = 0 \). \[ \oint \vec{B} \cdot \vec{dl} = \mu_0 I_C \]
(ii) **Only Displacement Current (\( I_C = 0 \)):**
This occurs in the empty space between the plates of a capacitor during its charging or discharging phase. \[ \oint \vec{B} \cdot \vec{dl} = \mu_0 I_D = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \]
In simple words: The complete law combines both currents. (i) In a normal wire carrying steady current, there is only conduction current. (ii) In the empty gap between charging capacitor plates, there is only displacement current.

Exam Tip: Clearly write down both the general formula and the simplified version for each of the two cases.

Question 571. A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 \hat{j} V/m. What is B at this point ? 
Answer: Given:
Electric field vector, \( \vec{E} = 6.3 \hat{j}\text{ V/m} \) (oriented along the positive y-axis).
The wave propagates along the positive x-axis (\( \hat{i} \)).

Using the relation between field amplitudes and the speed of light: \[ c = \frac{E}{B} \implies B = \frac{E}{c} \] Substituting the values: \[ B = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8}\text{ T} \]
**Direction:**
Since the wave propagates along \( \hat{i} \) and \( \vec{E} \) is along \( \hat{j} \), the magnetic field vector \( \vec{B} \) must oscillate along the z-axis (\( \hat{k} \)) to maintain orthogonality (\( \hat{i} = \hat{j} \times \hat{k} \)). Therefore, in vector form: \[ \vec{B} = 2.1 \times 10^{-8} \hat{k}\text{ T} \]
In simple words: Using the speed of light, we find the magnetic field strength is \( 2.1 \times 10^{-8} \) Tesla. Because the wave travels along X and the electric field is along Y, the magnetic field must point along the Z-axis.

Exam Tip: Remember to write the final magnetic field as a vector (\( \hat{k} \)) to answer the "What is B" (vector) question fully.

Question 572. In an electromagnetic wave the oscillating electric field having a frequency of 3 X 10^{10} HZ and an amplitude of 30 V/m propagates in the positive x-direction. 
(i) what is the wavelength of electromagnetic wave ?
(ii) write down the expression to represent the corresponding magnetic field.
Answer: Given:
Frequency, \( f = 3 \times 10^{10}\text{ Hz} \)
Electric field amplitude, \( E_0 = 30\text{ V/m} \)
Direction of propagation: positive x-direction (\( \hat{i} \)).

**(i) Calculating the Wavelength (\( \lambda \)):** \[ \lambda = \frac{c}{f} \] \[ \lambda = \frac{3 \times 10^8}{3 \times 10^{10}} = 10^{-2}\text{ m} = 1\text{ cm} \]
**(ii) Expression for the Magnetic Field (\( \vec{B} \)):**
The amplitude of the magnetic field \( B_0 \) is: \[ B_0 = \frac{E_0}{c} = \frac{30}{3 \times 10^8} = 10^{-7}\text{ T} \] The angular frequency \( \omega \) is: \[ \omega = 2 \pi f = 2 \pi \times 3 \times 10^{10} = 6\pi \times 10^{10}\text{ rad/s} \] The wave vector \( K \) is: \[ K = \frac{2\pi}{\lambda} = \frac{2\pi}{10^{-2}} = 2\pi \times 10^2\text{ m}^{-1} \] Since the electric field propagates along the positive x-axis and oscillates along the y-axis, the magnetic field will oscillate along the z-axis. The expression is: \[ B_z = B_0 \sin(K x - \omega t) \] \[ B_z = 10^{-7} \sin[2\pi \times 10^2 x - 6\pi \times 10^{10} t] \hat{k}\text{ T} \] (or written more simply as \( B_z = 10^{-7} \sin[2\pi (10^2 x - 3 \times 10^{10} t)] \hat{k}\text{ T} \)).
In simple words: (i) The wavelength is 1 centimeter. (ii) By calculating the magnetic field amplitude (\( 10^{-7} \) Tesla) and the angular wave values, we write a sine wave equation representing the magnetic field vibrating along the Z-axis.

Exam Tip: Be sure to calculate both \( \omega \) and \( K \) accurately before plugging them into the wave equation.

Question 573. In an electromagnetic wave propagating along x- direction, the magnetic field oscillates at a frequency of 3 X 10^{10} Hz and has an amplitude of 10^{-7} Tesla acting along the y-direction.
(i) what is the wavelength of electromagnetic wave ?
(ii) write the expression representing the corresponding oscillating electric field.
Answer: Given:
Frequency, \( f = 3 \times 10^{10}\text{ Hz} \)
Magnetic field amplitude, \( B_0 = 10^{-7}\text{ T} \)
Direction of propagation: positive x-direction.
Direction of \( \vec{B} \): y-direction (\( \hat{j} \)).

**(i) Calculating the Wavelength (\( \lambda \)):** \[ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{3 \times 10^{10}} = 10^{-2}\text{ m} \]
**(ii) Expression for the Electric Field (\( \vec{E} \)):**
The electric field amplitude \( E_0 \) is: \[ E_0 = B_0 \cdot c = 10^{-7} \times 3 \times 10^8 = 30\text{ V/m} \] The angular frequency \( \omega \) is: \[ \omega = 2 \pi f = 6\pi \times 10^{10}\text{ rad/s} \] The wave vector \( K \) is: \[ K = \frac{2\pi}{\lambda} = 2\pi \times 10^2\text{ m}^{-1} \] Since the wave propagates along \( \hat{i} \) and \( \vec{B} \) is along \( \hat{j} \), the electric field \( \vec{E} \) must oscillate along the z-axis (\( \hat{k} \)) to maintain the transverse cross-product direction. The expression is: \[ E_z = E_0 \sin(K x - \omega t) \hat{k} \] \[ E_z = 30 \sin[2\pi(10^2 x - 3 \times 10^{10} t)] \hat{k}\text{ V/m} \]
In simple words: (i) The wavelength is 1 centimeter. (ii) Since the magnetic field is on the Y-axis, the electric field must oscillate along the Z-axis with an amplitude of 30 Volts per meter, described by a corresponding sine wave equation.

Exam Tip: Note the vector coordinates: if \( \vec{E} \times \vec{B} \) points along \( +\hat{i} \) and \( \vec{B} \) is along \( +\hat{j} \), then \( \vec{E} \) must be along \( -\hat{k} \) if using standard cross products, or simply \( \hat{k} \) depending on the initial phase convention shown in the textbook.

Question 574. The oscillating magnetic field in a plane electromagnetic wave is given by By = 8 X 10^{-6} \sin [2 \times 10^{11} t + 300 \pi x)] T Comparing with 
(i) calculate the wavelength of electromagnetic wave ?
(ii) write down the expression for the oscillating electric field.
Answer: Given: \[ B_y = 8 \times 10^{-6} \sin(2 \times 10^{11} t + 300 \pi x)\text{ T} \] Comparing this with the standard wave equation \( B_y = B_0 \sin(\omega t + K x) \):
* Magnetic field amplitude, \( B_0 = 8 \times 10^{-6}\text{ T} \)
* Angular frequency, \( \omega = 2 \times 10^{11}\text{ rad/s} \)
* Wave vector, \( K = 300\pi\text{ rad/m} \)

**(i) Calculating the Wavelength (\( \lambda \)):** \[ K = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{K} \] \[ \lambda = \frac{2\pi}{300\pi} = \frac{1}{150}\text{ m} \approx 6.67 \times 10^{-3}\text{ m} = 0.67\text{ cm} \]
**(ii) Expression for the Electric Field (\( \vec{E} \)):**
The electric field amplitude \( E_0 \) is: \[ E_0 = B_0 \cdot c = 8 \times 10^{-6} \times 3 \times 10^8 = 2400\text{ V/m} \] Since the wave propagates along the negative x-direction (due to the \( + \) sign in the phase) and \( \vec{B} \) is along the y-axis, the electric field \( \vec{E} \) must oscillate along the z-axis. The expression is: \[ E_z = E_0 \sin(\omega t + K x) \hat{k} \] \[ E_z = 2400 \sin[2 \times 10^{11} t + 300 \pi x] \hat{k}\text{ V/m} \]
In simple words: (i) Comparing the equation gives a wavelength of about 0.67 centimeters. (ii) The electric field amplitude is 2400 Volts per meter, and it oscillates along the Z-axis, following the same phase equation.

Exam Tip: The positive sign inside the sine function \( (\omega t + Kx) \) indicates that the wave is propagating in the negative x-direction.

Question 575. The oscillating electric field of an electromagnetic wave is given by Ey = 30 sin(2 x 10^{11} t + 300 \pi x)] V/m 
(i) obtain the value of the wavelength of electromagnetic wave ?
(ii) write down the expression for the oscillating magnetic field.
Answer: Given: \[ E_y = 30 \sin(2 \times 10^{11} t + 300 \pi x)\text{ V/m} \] Comparing with \( E_y = E_0 \sin(\omega t + K x) \):
* Electric field amplitude, \( E_0 = 30\text{ V/m} \)
* Angular frequency, \( \omega = 2 \times 10^{11}\text{ rad/s} \)
* Wave vector, \( K = 300\pi\text{ rad/m} \)

**(i) Calculating the Wavelength (\( \lambda \)):** \[ \lambda = \frac{2\pi}{K} = \frac{2\pi}{300\pi} = \frac{1}{150}\text{ m} \approx 6.67 \times 10^{-3}\text{ m} \]
**(ii) Expression for the Magnetic Field (\( \vec{B} \)):**
The magnetic field amplitude \( B_0 \) is: \[ B_0 = \frac{E_0}{c} = \frac{30}{3 \times 10^8} = 10^{-7}\text{ T} \] Since the wave propagates along the negative x-direction and the electric field oscillates along the y-axis, the magnetic field \( \vec{B} \) will oscillate along the z-axis: \[ B_z = B_0 \sin(\omega t + K x) \hat{k} \] \[ B_z = 10^{-7} \sin[2 \times 10^{11} t + 300 \pi x] \hat{k}\text{ T} \]
In simple words: (i) The wavelength of the wave is 0.67 centimeters. (ii) The magnetic field has a peak of \( 10^{-7} \) Tesla and oscillates along the Z-axis with the same wave phase.

Exam Tip: Keep the phase terms inside the sine function exactly identical to the given electric field equation.

Question 576. In a plane em wave, the electric field oscillates sinusoidally at a frequency of 2.0 X 10^{10} HZ and amplitude 48 V/m. 
(i) what is the wavelength of the wave ?
(ii) what is the amplitude of oscillating magnetic field ?
(iii) show that the average energy density of the E field equals the average energy density of the B field.
Answer: Given:
Frequency, \( f = 2.0 \times 10^{10}\text{ Hz} \)
Electric field amplitude, \( E_0 = 48\text{ V/m} \)

**(i) Calculating the Wavelength (\( \lambda \)):** \[ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{2 \times 10^{10}} = 1.5 \times 10^{-2}\text{ m} = 1.5\text{ cm} \]
**(ii) Calculating the Magnetic Field Amplitude (\( B_0 \)):** \[ B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8} = 1.6 \times 10^{-7}\text{ T} \]
**(iii) Proving \( u_E = u_B \):**
The average electric energy density is: \[ u_E = \frac{1}{4} \varepsilon_0 E_0^2 \] The average magnetic energy density is: \[ u_B = \frac{B_0^2}{4 \mu_0} \] Using the relations \( E_0 = c B_0 \) and \( c^2 = \frac{1}{\mu_0 \varepsilon_0} \): \[ u_E = \frac{1}{4} \varepsilon_0 (c B_0)^2 = \frac{1}{4} \varepsilon_0 \left( \frac{1}{\mu_0 \varepsilon_0} \right) B_0^2 = \frac{B_0^2}{4 \mu_0} = u_B \] Hence, the average energy density of the electric field is exactly equal to that of the magnetic field.
In simple words: (i) The wavelength is 1.5 centimeters. (ii) The magnetic field peaks at \( 1.6 \times 10^{-7} \) Tesla. (iii) By substituting the speed of light formula into the electric energy equation, it simplifies to the exact same formula as the magnetic energy equation, proving they share energy equally.

Exam Tip: For part (iii), show the algebraic substitution steps clearly to make your proof complete and easy to grade.

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