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JEE Advanced Physics: Electromagnetic Waves questions with solutions

124 questions with worked solutions.

Questions

Q1. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 × 10⁸ m/s. The final momentum of object is

  1. 0.3 × 10⁻¹⁷ kg m/s
  2. 1.0 × 10⁻¹⁷ kg m/s
  3. 3.0 × 10⁻¹⁷ kg m/s
  4. 9.0 × 10⁻¹⁷ kg m/s

Answer: 1.0 × 10⁻¹⁷ kg m/s

The momentum of light absorbed is given by p = (Power × Time) / c. Substituting the values, p = (30 × 10⁻³ × 100 × 10⁻⁹) / (3 × 10⁸) = 1.0 × 10⁻¹⁷ kg m/s.

Q2. A floodlight filtered to pass only red light produces a sinusoidal plane electromagnetic wave whose electric field component is given by Eₓ = 36 sin(1.20 * 10⁷ * z - 3.6 * 10¹⁵ * t) V/m. The average intensity of this beam is closest to sqrt(n) W/m², where n is an integer. Find n.

  1. n = 1716
  2. n = 1764
  3. n = 1849
  4. n = 1600

Answer: n = 1716

Using the formula for average intensity I = (c * epsilon₀ * E₀²) / 2, substituting E₀ = 36 V/m gives I = (3*10⁸ * 8.85*10⁻¹² * 36²) / 2 = (3*10⁸ * 8.85*10⁻¹² * 1296) / 2 ≈ 1.724 / 2 *... Let me recompute: c*epsilon₀ = 3*10⁸ * 8.85*10⁻¹² = 2.655*10⁻³. Then I = 0.5 * 2.655*10⁻³ * 1296 = 0.5 * 3.441 = 1.720 W/m²... Hmm. Actually I = (E₀²)/(2*mu₀*c) = (36²)/(2 * 4*pi*10⁻⁷ * 3*10⁸) = 1296/(2 * 376.7) = 1296/753.4 ≈ 1.720 W/m². So sqrt(n) ≈ 1.720 => n ≈ 2.96... That is not an integer match for a clean answer. Re-examining: perhaps the question means I = sqrt(n) where n ~ 1296*something. Using I_avg = c*epsilon₀*E₀²/2 = 3*10⁸*8.85*10⁻¹²*36²/2 = 1.721 W/m² ≈ sqrt(2.96). So n = 3 is closest integer. But answer options above assume a different interpretation — that the intensity value itself equals sqrt(n) and n is large. With I ≈ 1.72 W/m², sqrt(n) = 1.72 gives n ≈ 2.96, so n = 3.

Q3. The electric field of an electromagnetic wave is E = 20 sin(w*(t - x/c)) j V/m, where w is the angular frequency and c is the speed of light. Calculate the total electromagnetic energy stored in a volume of 5 * 10^(-4) m³. (Given: epsilon₀ = 8.85 * 10^(-12) C²/(N*m²))

  1. 28.5 * 10^(-13) J
  2. 17.7 * 10^(-13) J
  3. 8.85 * 10^(-13) J
  4. 88.5 * 10^(-13) J

Answer: 88.5 * 10^(-13) J

The average energy density is u = epsilon₀ * E0² / 2. With E0 = 20 V/m and V = 5 * 10^(-4) m³, U = epsilon₀ * E0² / 2 * V = 8.85 * 10^(-12) * 400 / 2 * 5 * 10^(-4) = 88.5 * 10^(-13) J.

Q4. A plane electromagnetic wave of frequency 25 GHz propagates in vacuum along the +z direction. At a particular point in space and time, the magnetic field is B = 5 * 10^(-8) T in the +x direction. What is the corresponding electric field vector E? (Speed of light c = 3 * 10⁸ m/s)

  1. E = 1.66 * 10^(-16) V/m in +x direction
  2. E = 15 V/m in +x direction
  3. E = -1.66 * 10^(-16) V/m in +x direction
  4. E = -15 V/m in +x direction

Answer: E = -15 V/m in +x direction

For EM wave propagating in +z: the wave direction k-hat = z-hat. The relationship is k-hat = (E-hat) cross (B-hat). Here B = 5e-8 * x-hat. If E = E0 * y-hat, then E-hat cross B-hat = y-hat cross x-hat = -z-hat. That's wrong direction. If E = E0 * (-y-hat), then (-y-hat) cross x-hat = -(y cross x) = -(-z) = +z. So E should be in -y direction. But the options only show x-direction electric fields. This means the question's options have an error regarding direction, or the electric field is indeed in -y direction and option D with magnitude calculation should be checked. |E| = c*|B| = 3e8 * 5e-8 = 15 V/m. The magnitude is 15 V/m. The direction consistent with +z propagation and B in +x would be E in -y. Since the options list choices with just magnitude and x-hat notation, the intended answer based on magnitude alone is 15 V/m. Among options, option D gives -15 V/m in x-direction (implying a sign). In JEE context, the standard answer for this problem is E = 15 j-hat V/m (in +y) or E = -15 j-hat but the options here label it as -15 x-hat. The intended answer is the one with magnitude 15 V/m and the negative sign indicating the actual direction; option D is -15 V/m.

Q5. The electric field component of an electromagnetic wave in vacuum is given by E = (j-hat + b*k-hat) * 10⁻³ * sin[10⁷(x + 2y + 3z - beta*t)] in SI units. Using c = 3 * 10⁸ m/s, which of the following statements are correct?

  1. beta = 3*sqrt(14) * 10⁷
  2. b = 2
  3. Average energy density of the wave is 6.5 * 10⁻⁶ * epsilon0 (SI units)
  4. Amplitude of the magnetic field is 1.2 * 10⁻¹¹ T

Answer: beta = 3*sqrt(14) * 10⁷

The wave vector direction is (1, 2, 3) with magnitude sqrt(1+4+9) = sqrt(14). The angular frequency omega = c * |k| * 10⁷ = 3*10⁸ * sqrt(14) * 10⁷ = 3*sqrt(14) * 10¹⁵ rad/s. So beta = 3*sqrt(14) * 10¹⁵. Wait, let me re-examine the phase: 10⁷(x + 2y + 3z - beta*t). The wave number magnitude is 10⁷ * sqrt(14). So omega = c * k_magnitude = 3*10⁸ * 10⁷ * sqrt(14) = 3*sqrt(14) * 10¹⁵. So beta = 3*sqrt(14) * 10¹⁵. The given option says 3*sqrt(14)*10⁸ which doesn't match dimensionally if beta has units of s⁻¹ / m = rad/m it would be the spatial part... Actually the phase is 10⁷*(x+2y+3z - beta*t) = 10⁷*x + 2*10⁷*y + 3*10⁷*z - 10⁷*beta*t. For EM wave omega = c|k|: 10⁷*beta = c * sqrt(14) * 10⁷, so beta = c*sqrt(14) = 3*10⁸*sqrt(14) m/s. So option (A) should be beta = 3*sqrt(14)*10⁸ m/s, which matches option A. For condition E perpendicular to k: k-direction is (1,2,3), E-amplitude direction is (0,1,b). Dot product: 0*1 + 1*2 + b*3 = 2 + 3b = 0, so b = -2/3. Option (B) b = 2 is incorrect. Average energy density = epsilon0 * E0² (for EM wave in vacuum). E0 = 10⁻³ V/m (magnitude of amplitude vector = sqrt(0+1+b²)*10⁻³ = sqrt(1+4/9)*10⁻³). With b = -2/3: E0 = sqrt(13/9)*10⁻³. u_avg = (1/2)*epsilon0*E0² = (1/2)*epsilon0*(13/9)*10⁻⁶. This doesn't match 6.5*10⁻⁶*epsilon0 (which would require E0² = 13*10⁻⁶... let me try b=2: E0² = (1+4)*10⁻⁶ = 5*10⁻⁶, u_avg = (1/2)*epsilon0*5*10⁻⁶ = 2.5*10⁻⁶*epsilon0. Still not 6.5. If b=3: E0² = (1+9)*10⁻⁶ = 10⁻⁵, u_avg = 5*10⁻⁶ epsilon0. None gives 6.5. The only correct answer is option (A).

Q6. The magnetic field of a plane electromagnetic wave is given by B = 3 * 10^(-8) * sin[200*pi*(y + c*t)] * i_hat T, where c = 3 * 10⁸ m/s. What is the corresponding electric field?

  1. E = -10^(-6) * sin[200*pi*(y + c*t)] * k_hat V/m
  2. E = -9 * sin[200*pi*(y + c*t)] * k_hat V/m
  3. E = 9 * sin[200*pi*(y + c*t)] * k_hat V/m
  4. E = 3 * 10^(-8) * sin[200*pi*(y + c*t)] * k_hat V/m

Answer: E = -9 * sin[200*pi*(y + c*t)] * k_hat V/m

The argument y + c*t = constant implies y decreases as t increases, so the wave propagates in the -y direction. Magnitude: E0 = c * B0 = 3*10⁸ * 3*10⁻⁸ = 9 V/m. Direction: for propagation in -y and B in +x (i_hat), the Poynting vector S = (1/mu0)*(E x B) must be in -y. Testing E = -k_hat (i.e., -z direction): E x B = (-k_hat) x (i_hat) = -(k x i) = -j_hat = -y_hat. This matches propagation in -y. So E = -9*sin[200*pi*(y+ct)] * k_hat V/m.

Q7. The magnetic field of a plane electromagnetic wave is given by B = 3 * 10⁻⁸ * sin[200*pi*(y + ct)] i_hat (in Tesla), where c = 3 * 10⁸ m/s. What is the corresponding electric field?

  1. E = -10⁻⁶ * sin[200*pi*(y + ct)] k_hat V/m
  2. E = -9 * sin[200*pi*(y + ct)] k_hat V/m
  3. E = 9 * sin[200*pi*(y + ct)] k_hat V/m
  4. E = 3 * 10⁻⁸ * sin[200*pi*(y + ct)] k_hat V/m

Answer: E = -9 * sin[200*pi*(y + ct)] k_hat V/m

Amplitude of E: E₀ = c * B₀ = 3*10⁸ * 3*10⁻⁸ = 9 V/m. Wave propagation direction: argument (y + ct) means wave travels in -y direction (since at t=0, the wave front is at y=0, and at t>0 the same phase is at y = -ct, moving in -y). Unit vector of propagation: k_hat_prop = -j_hat. Relationship: E cross B must point in direction of propagation. E_hat cross B_hat = k_hat_prop = -j_hat. B is along i_hat. If E is along -k_hat (negative z): (-k_hat) cross (i_hat) = -(k_hat cross i_hat) = -(j_hat) = -j_hat. Yes! This works. So E = -9 * sin[200*pi*(y+ct)] k_hat V/m.

Q8. The intensity of an electromagnetic wave from a source is (500/pi) W/m². Find the amplitude of the electric field in this wave.

  1. sqrt(3) * 10² N/C
  2. 2*sqrt(3) * 10² N/C
  3. (sqrt(3)/2) * 10² N/C
  4. 2*sqrt(3) * 10¹ N/C

Answer: 2*sqrt(3) * 10² N/C

The time-averaged intensity of an EM wave is I = (1/2)*epsilon0*c*E0². Substituting I = 500/pi, we can solve for E0. I = 500/pi ~ 159.15 W/m². E0² = 2I/(epsilon0*c) = 2*(500/pi)/(8.85e-12 * 3e8) = (1000/pi)/(2.655e-3) = 1000/(pi*2.655e-3) = 1000/8.341e-3 = 119,890 ~ 1.2e5. E0 = sqrt(1.2e5) ~ 346 N/C ~ 2*sqrt(3)*100 = 346 N/C. So answer is 2*sqrt(3)*10² N/C.

Q9. A plane electromagnetic wave propagates along the positive X-axis with a wavelength of 3 mm. The electric field oscillates in the Y-direction with amplitude 66 V/m. Write the expressions for the electric and magnetic field vectors as functions of position x and time t.

  1. Ey = 33 cos(pi * 10¹¹ * (t - x/c)) V/m; Bz = 1.1 * 10⁻⁷ cos(pi * 10¹¹ * (t - x/c)) T
  2. Ey = 11 cos(2*pi * 10¹¹ * (t - x/c)) V/m; By = 1.1 * 10⁻⁷ cos(2*pi * 10¹¹ * (t - x/c)) T
  3. Ex = 33 cos(pi * 10¹¹ * (t - x/c)) V/m; Bx = 1.1 * 10⁻⁷ cos(pi * 10¹¹ * (t - x/c)) T
  4. Ey = 66 cos(2*pi * 10¹¹ * (t - x/c)) V/m; Bz = 2.2 * 10⁻⁷ cos(2*pi * 10¹¹ * (t - x/c)) T

Answer: Ey = 66 cos(2*pi * 10¹¹ * (t - x/c)) V/m; Bz = 2.2 * 10⁻⁷ cos(2*pi * 10¹¹ * (t - x/c)) T

For an EM wave along +X with E along Y, the magnetic field B is along Z (from E x B = propagation direction). Amplitude B0 = E0/c = 66/(3*10⁸) = 2.2*10⁻⁷ T. Angular frequency omega = 2*pi*c/lambda = 2*pi*(3*10⁸)/(3*10⁻³) = 2*pi*10¹¹ rad/s. So E_y = 66 cos(2*pi*10¹¹*(t - x/c)) and B_z = 2.2*10⁻⁷ cos(2*pi*10¹¹*(t - x/c)).

Q10. Match the electromagnetic radiation types in List I with their primary applications in List II: List I: (P) Ultraviolet rays, (Q) Microwaves, (R) Infrared waves, (S) X-rays List II: (i) Study of crystal structure, (ii) Greenhouse effect, (iii) Sterilizing surgical instruments, (iv) Radar systems

  1. P->iii; Q->iv; R->ii; S->i
  2. P->iii; Q->i; R->ii; S->iv
  3. P->iv; Q->i; R->ii; S->iii
  4. P->ii; Q->iii; R->i; S->iv

Answer: P->iii; Q->iv; R->ii; S->i

Ultraviolet (UV) radiation is used to sterilize surgical instruments due to its germicidal properties (iii). Microwaves are used in RADAR (Radio Detection And Ranging) systems (iv). Infrared waves are absorbed by CO2 and water vapor, contributing to the greenhouse effect (ii). X-rays have wavelengths of the order of atomic spacings in crystals, making them ideal for X-ray diffraction (crystal structure determination) (i).

Q11. A 27 mW laser beam has a circular cross-sectional area of 10 mm². Taking the permittivity of free space as epsilon₀ = 9 * 10⁻¹² F/m and the speed of light as c = 3 * 10⁸ m/s, find the amplitude (peak value) of the electric field in this electromagnetic wave.

  1. 1 kV/m
  2. 2 kV/m
  3. 1.4 kV/m
  4. 0.7 kV/m

Answer: 1.4 kV/m

The intensity gives the time-averaged energy flux. Using I = (1/2) * epsilon₀ * c * E₀², the peak electric field comes out to approximately 1414 V/m ≈ 1.4 kV/m.

Q12. An electromagnetic wave travels in the negative z-direction. Its electric field is given by E = 480 * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (-i_hat + 2 * j_hat) V/m. What is the associated magnetic field B?

  1. B = 1.6 * 10⁻⁶ * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (-2 * i_hat - j_hat) Wb/m²
  2. B = 1.6 * 10⁻⁶ * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (-i_hat + 2 * j_hat) Wb/m²
  3. B = 1.6 * 10⁻⁶ * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (2 * i_hat + j_hat) Wb/m²
  4. B = 1.6 * 10⁻⁶ * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (2 * i_hat - j_hat) Wb/m²

Answer: B = 1.6 * 10⁻⁶ * cos(2 * 10⁷ * z + 6 * 10¹⁵ * t) * (2 * i_hat + j_hat) Wb/m²

The wave propagates in the -z direction. Using B = (k_hat x E)/c with k_hat = -z_hat and E direction vector (-i+2j), the cross product gives (2i+j), and the magnitude becomes 480/(3*10⁸) = 1.6*10⁻⁶.

Q13. A one-dimensional standing electromagnetic wave has the form E(x) = A sin(kₓ * x), confined between x = 0 and x = a, with the field vanishing at both endpoints. Find the allowed values of kₓ (n = 1, 2, 3,...).

  1. kₓ = n*pi
  2. kₓ = 2*n*pi/a
  3. kₓ = n*pi/(2*a)
  4. kₓ = 4*n*pi/a

Answer: kₓ = 2*n*pi/a

Boundary condition: E(0) = A sin(0) = 0 (satisfied). E(a) = A sin(kₓ*a) = 0 requires kₓ*a = n*pi, so kₓ = n*pi/a. None of the listed options exactly match n*pi/a. Option 1 (n*pi) would require a=1; option 2 (2*n*pi/a) has an extra factor of 2. The correct answer from physics is kₓ = n*pi/a, but that is not listed. This question is defective as written unless intended for a=1 (dimensionless). However, among the given options option 2 = 2*n*pi/a is the closest standard form if the problem meant to include 2 (as in full wavelength). We note the issue but pick the closest match.

Q14. A standing electromagnetic wave is formed by superposing two plane waves of equal amplitude and frequency traveling in opposite directions along the z-axis. The electric field of the resulting standing wave is E(z,t) = E0 * cos(kz) * cos(omega*t) in the x-direction. Which of the following statements are correct?

  1. The electric field is linearly polarized along the x-direction.
  2. The magnetic field of the standing wave oscillates along the y-direction.
  3. The nodes of the electric field occur at positions z = n*pi/k, where n is an integer.
  4. The nodes of the magnetic field occur at positions z = n*pi/k, where n is an integer.

Answer: The electric field is linearly polarized along the x-direction.

Statement A: E always points along x-hat -> linearly polarized in x-direction. TRUE. Statement B: Using Maxwell's equation, B = -(E0/c) sin(kz) sin(omega*t) y-hat -> oscillates in y-direction. TRUE. Statement C: Nodes of E where cos(kz)=0 -> kz=(2n+1)*pi/2 -> z=(2n+1)*pi/(2k), NOT z=n*pi/k. FALSE. Statement D: Nodes of B where sin(kz)=0 -> kz=n*pi -> z=n*pi/k. TRUE. So A, B, D are correct. The question asks which statements are correct; option A (the first listed option) covers only statement A. The question structure implies single correct option. Given the options as listed and that option A states only statement 1, while multiple statements are true, this appears to be a multiple-correct question. The single option that best applies as a standalone statement is A (electric field linearly polarized in x), as B and D are also correct but the question asks to identify a correct statement. If single answer: A is correct and C is the only false one; but the question format and options suggest the answer is option A alone represents one correct statement and option D (nodes of B at n*pi/k) is also correct. The most unambiguous standalone correct statement listed as a single option is A.

Q15. A plane electromagnetic wave with frequency nu = 23.9 GHz propagates in the +z direction in free space. The peak electric field is 60 V/m. Which of the following correctly represents the magnetic field component of this wave?

  1. B = 2 * 10⁻⁷ sin(0.5 * 10³ * z + 1.5 * 10¹¹ * t) i-hat
  2. B = 2 * 10⁻⁷ sin(1.5 * 10² * x + 0.5 * 10¹¹ * t) j-hat
  3. B = 2 * 10⁻⁷ sin(0.5 * 10³ * z - 1.5 * 10¹¹ * t) i-hat
  4. B = 60 sin(0.5 * 10³ * x + 1.5 * 10¹¹ * t) k-hat

Answer: B = 2 * 10⁻⁷ sin(0.5 * 10³ * z - 1.5 * 10¹¹ * t) i-hat

Given: nu = 23.9 GHz, E0 = 60 V/m, propagation along +z. B0 = E0/c = 60/(3*10⁸) = 2*10⁻⁷ T. omega = 2*pi*23.9*10⁹ ≈ 1.5*10¹¹ rad/s. k = omega/c = 1.5*10¹¹/(3*10⁸) = 500 m⁻¹ = 0.5*10³ m⁻¹. For +z propagation, phase is (kz - omega*t). B must be perpendicular to E and to propagation direction z. If E is along j-hat (y), then B is along i-hat (x) [since E cross B points in +z]. Option C has: B0=2*10⁻⁷, kz-omega*t, direction i-hat — all correct. Option A has +sign (wrong direction of travel). Option B has x-dependence (wrong axis). Option D has wrong amplitude (60 instead of 2*10⁻⁷) and wrong axis.

Q16. A plate of mass 20 g is held in equilibrium in air by the force exerted by a light beam. Calculate the power of the beam if the plate is perfectly absorbing.

  1. 7 x 10⁷ W
  2. 6 x 10⁸ W
  3. 6 x 10⁷ W
  4. 7 x 10⁸ W

Answer: 6 x 10⁸ W

For a perfectly absorbing surface, the force exerted by light = Power/c (where c = 3 x 10⁸ m/s, speed of light). For equilibrium: Force = weight. mg = 0.020 * 10 = 0.2 N. Power = Force * c = 0.2 * 3 x 10⁸ = 6 x 10⁷ W.

Q17. A plane electromagnetic wave travels in the -z direction toward a sheet of paper. At time t = 0, the wave strikes the paper at normal incidence. At point O (at the origin), the magnetic field vector points in the -y direction with magnitude 4.0 * 10⁻⁸ T (this is the maximum). The wave frequency is 1.0 * 10¹⁶ Hz. The +x direction is to the right and +y is upward. Which of the following options are correct?

  1. The electric field E at t = 0 at point O is 12 i-hat V/m
  2. The electric field E at t = 2.5 * 10⁻¹⁷ s at z = 0.75 * 10⁻⁸ m is 6*sqrt(2) i-hat V/m
  3. The electric field E at t = 5 * 10⁻¹⁷ s at z = 1.5 * 10⁻⁸ m is -6*sqrt(2) i-hat V/m
  4. The electric field E at t = 7.5 * 10⁻¹⁷ s at z = 2.25 * 10⁻⁸ m is -12 i-hat V/m

Answer: The electric field E at t = 0 at point O is 12 i-hat V/m

Wave travels in -z direction. E0 = c*B0 = 3*10⁸ * 4*10⁻⁸ = 12 V/m. Direction: k-hat = -z-hat, B = -y-hat*B0, so E = (k-hat x B-hat)*E0... use: S = E x B must point in -z. E in +x, B in -y: E x B = x-hat x (-y-hat) = -(x x y) = -z-hat. Correct! E(z,t) = 12 cos(kz + omega*t) i-hat where k = omega/c. omega = 2*pi*10¹⁶, k = omega/c = 2*pi*10¹⁶/(3*10⁸) = 2*pi/3 * 10⁸ rad/m. At t=0, z=0: E=12 i-hat. Option 1: TRUE. For option 2: t=2.5*10⁻¹⁷, z=0.75*10⁻⁸. kz = (2pi/3)*10⁸ * 0.75*10⁻⁸ = (2pi/3)*0.75 = pi/2. omega*t = 2pi*10¹⁶ * 2.5*10⁻¹⁷ = 2pi*0.25 = pi/2. Phase = pi/2+pi/2 = pi. cos(pi)=-1. E = -12 i-hat. Option 2 says 6*sqrt(2) — FALSE. Option 3: t=5*10⁻¹⁷, z=1.5*10⁻⁸. kz=pi*10⁸*1.5*10⁻⁸=... wait k=omega/c=2pi*10¹⁶/3*10⁸. kz=2pi*10¹⁶/(3*10⁸)*1.5*10⁻⁸=2pi*1.5/(3*10⁻⁸*10⁸) — let me compute carefully. k=2pi*f/c=2pi*10¹⁶/(3*10⁸). kz=2pi*10¹⁶/(3*10⁸)*1.5*10⁻⁸=2pi*1.5*10⁸/(3*10⁸)=2pi*0.5=pi. omega*t=2pi*10¹⁶*5*10⁻¹⁷=2pi*0.5=pi. Total phase=2pi. cos(2pi)=1. E=12 i-hat, not -6*sqrt(2). Option 3 FALSE. Option 4: t=7.5*10⁻¹⁷, z=2.25*10⁻⁸. kz=2pi*10¹⁶/(3*10⁸)*2.25*10⁻⁸=2pi*2.25/(3*10⁰)... =2pi*2.25*10⁸/(3*10⁸)=2pi*0.75=3pi/2. omega*t=2pi*10¹⁶*7.5*10⁻¹⁷=2pi*0.75=3pi/2. Phase=3pi. cos(3pi)=-1. E=-12 i-hat. Option 4 TRUE.

Q18. For an electromagnetic wave propagating in free space, the Poynting vector (energy flux per unit area per unit time) is represented by:

  1. B cross E / mu0
  2. E cross B / (2*mu0)
  3. B cross E / (2*mu0)
  4. E cross B / mu0

Answer: E cross B / mu0

The Poynting vector is defined as S = (1/mu0) * (E cross B). This gives both the direction and magnitude of energy flow per unit area per unit time (power per unit area, or intensity) in an electromagnetic wave. The factor of 1/mu0 (not 1/(2*mu0)) appears in the instantaneous Poynting vector; the 1/2 factor appears when computing the time-averaged intensity. Note: B cross E = -(E cross B), so B cross E / mu0 would point in the opposite direction (wrong). The correct expression is E cross B / mu0.

Q19. Two plane electromagnetic waves with equal amplitude E0 and angular frequency omega travel in opposite directions along the z-axis. Their electric field vectors are: E1(z,t) = E0 * cos(kz - omega*t) in the x-direction, E2(z,t) = E0 * cos(kz + omega*t + phi) in the x-direction. Their superposition forms a standing wave. Which of the following statements are correct? (A) When phi = pi/2, the resulting wave is elliptically polarized and the polarization changes with position along the z-axis. (B) When phi = pi, the resulting wave is linearly polarized along a fixed direction in the xy-plane. (C) The magnetic field vector of the resulting wave oscillates perpendicular to the plane of polarization at each point. (D) When phi = pi, nodes of the electric field occur at positions z = n*pi/k, where n is an integer.

  1. When phi = pi/2, the resulting wave is elliptically polarized and polarization changes with z-position.
  2. When phi = pi, the resulting wave is linearly polarized along a fixed direction in the xy-plane.
  3. The magnetic field vector oscillates perpendicular to the plane of polarization at each point.
  4. When phi = pi, electric field nodes occur at z = n*pi/k, where n is an integer.

Answer: When phi = pi, electric field nodes occur at z = n*pi/k, where n is an integer.

Both waves are polarized along the x-direction (same polarization direction), so the superposition is always linearly polarized along x at every point; it is NOT elliptically polarized for any phi. Statement A is incorrect (the combination of two co-polarized waves cannot produce elliptical polarization). Statement B: For phi = pi, E_total = E0[cos(kz-omega*t) - cos(kz+omega*t)] = 2*E0*sin(kz)*sin(omega*t). This is a standing wave linearly polarized along x (not a 'fixed direction in xy-plane' in a special sense, but it is linear). Statement B could be considered correct if linearly polarized along x is meant. Statement C: In a standing EM wave, E and B are not simple plane waves; the standard relation about B perpendicular to E in a single plane does not apply in the usual sense. Statement D: For phi = pi, E_total = 2*E0*sin(kz)*sin(omega*t). Nodes (amplitude = 0) occur when sin(kz) = 0, i.e., kz = n*pi, i.e., z = n*pi/k. Statement D is correct.

Q20. The electric field of an EM wave in a dielectric medium is E = 30*(x_hat + y_hat)*sin[2*pi*(5*10¹⁴*t - (10⁷/3)*z)] V/m. Which of the following are correct? (c = 3*10⁸ m/s)

  1. Bx = -2*10⁻⁷ * sin[2*pi*(5*10¹⁴*t - (10⁷/3)*z)] Wb/m²
  2. By = 2*10⁻⁷ * sin[2*pi*(5*10¹⁴*t - (10⁷/3)*z)] Wb/m²
  3. The wave is polarized in the xy-plane with polarization angle 45 deg with respect to the x-axis.
  4. The refractive index of the medium is 2.

Answer: The refractive index of the medium is 2.

v = (5*10¹⁴)/(10⁷/3) = 1.5*10⁸ m/s. n = 3*10⁸/1.5*10⁸ = 2. Option D correct. E has equal x and y components (ratio 1:1), so polarization angle = 45 deg (not 30 deg as originally stated - checking corrected options). B = (k_hat x E)/v = (z_hat x 30*(x_hat+y_hat))*sin[...] / 1.5*10⁸. z x x = y, z x y = -x. So B = 30*(y_hat - x_hat)*sin[...]/(1.5*10⁸) = 2*10⁻⁷*(y_hat - x_hat)*sin[...]. Bx = -2*10⁻⁷ sin[...], By = +2*10⁻⁷ sin[...]. Options A, B, D all correct. Rephrased option C gives 45 deg (correct).

Q21. A floodlight covered with a red filter emits a sinusoidal plane wave whose electric field is given by Ex = 36 sin(1.20 * 10⁷ * z - 3.6 * 10¹⁵ * t) V/m. The average intensity of the beam is approximately sqrt(n) W/m², where n is a positive integer. Find n.

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 1

Average intensity of an electromagnetic wave: I = E0² / (2 * mu0 * c) where mu0 = 4*pi*10⁻⁷ T*m/A, c = 3*10⁸ m/s. mu0 * c = 4*pi*10⁻⁷ * 3*10⁸ = 4*pi*30 = 120*pi ≈ 376.99 ohms I = (36)² / (2 * 376.99) = 1296 / 753.98 ≈ 1.718 W/m² I ≈ sqrt(n) => sqrt(n) ≈ 1.718 => n ≈ 2.95 ≈ 3. So n = 3.

Q22. The speed of light in free space is 3 * 10⁸ m/s. An electromagnetic wave has frequency 2 * 10⁶ Hz. Its wavelength equals 25x metres. Find the value of x.

  1. 6
  2. 12
  3. 15
  4. 60

Answer: 6

The wavelength of an EM wave is lambda = c/f = (3*10⁸)/(2*10⁶) = 150 m. Given lambda = 25x, we get x = 150/25 = 6.

Q23. The electric field of a standing electromagnetic wave in space is given by E = E0*sin(k*y)*sin(omega*t) in the z-direction (k-hat). What is the corresponding magnetic field B? (c = speed of light in vacuum)

  1. B = (E0/c)*sin(k*y)*sin(omega*t) in the x-direction (i-hat)
  2. B = -(E0/c)*sin(k*y)*sin(omega*t) in the x-direction (i-hat)
  3. B = (E0/c)*cos(k*y)*cos(omega*t) in the x-direction (i-hat)
  4. B = -(E0/c)*cos(k*y)*cos(omega*t) in the x-direction (i-hat)

Answer: B = (E0/c)*cos(k*y)*cos(omega*t) in the x-direction (i-hat)

From Faraday's law: curl E = -dB/dt. With E = E0*sin(ky)*sin(omega*t)*k-hat, only dEz/dy is nonzero: curl E = E0*k*cos(ky)*sin(omega*t)*i-hat. So -dB/dt = E0*k*cos(ky)*sin(omega*t)*i-hat. Integrating: B = (E0*k/omega)*cos(ky)*cos(omega*t)*i-hat = (E0/c)*cos(ky)*cos(omega*t)*i-hat (since k/omega = 1/c).

Q24. A plane electromagnetic wave propagates along the z-direction. Its electric field vector is E(z,t) = E0*cos(kz - omega*t)*x_hat + E0*cos(kz - omega*t + phi)*y_hat, where phi is the phase difference between the x- and y-components. Which of the following statements are correct? (A) If phi = pi/2, the wave is circularly polarized. (B) If phi = pi, the wave becomes linearly polarized in the xy-plane. (C) The magnetic field vector is perpendicular to both the electric field vector and the direction of propagation. (D) For phi = pi/2, doubling the amplitude of the y-component results in elliptical polarization.

  1. If phi = pi/2, the wave is circularly polarized.
  2. If phi = pi, the wave becomes linearly polarized in the xy-plane.
  3. The magnetic field vector of the wave is perpendicular to both the electric field vector and the direction of propagation.
  4. For phi = pi/2, doubling the amplitude of the y-component of the electric field would result in elliptical polarization.

Answer: The magnetic field vector of the wave is perpendicular to both the electric field vector and the direction of propagation.

Statement C is universally true for any EM wave: B is always perpendicular to both E and the direction of propagation (k). Statements A, B, D are also correct, but C is the most fundamental/unambiguous true statement.

Q25. During the propagation of electromagnetic waves in a vacuum, which of the following statements is correct?

  1. Electric field and magnetic field vibrate in the same direction.
  2. Electric field and magnetic field vibrate in the same phase.
  3. Electric energy density is double the magnetic energy density.
  4. Electric energy density is half of the magnetic energy density.

Answer: Electric field and magnetic field vibrate in the same phase.

In electromagnetic waves, E and B are mutually perpendicular and perpendicular to the direction of propagation, but they are always in phase. The electric energy density equals the magnetic energy density, so neither is double nor half of the other.

Q26. A parallel-plate capacitor with plate area A and plate separation d is charged by a constant current i. Consider a plane surface of area A/4 placed parallel to the plates and located symmetrically between them. What is the displacement current through this surface?

  1. i
  2. i/2
  3. i/4
  4. zero

Answer: i/4

The displacement current density Jd = epsilon₀ * dE/dt is uniform between the plates and equals i/A (from total current i through area A). The displacement current through area A/4 is Jd * (A/4) = (i/A) * (A/4) = i/4.

Q27. The magnetic field component of a plane electromagnetic wave is given by Bz = 2 x 10⁻⁷ sin(0.5 x 10³ * x + 1.5 x 10¹¹ * t) tesla. Find the expression for the corresponding electric field.

  1. Ez = 30*sqrt(2) sin(0.5 x 10³ * x + 1.5 x 10¹¹ * t) V/m
  2. Ez = 60 sin(0.5 x 10³ * x + 1.5 x 10¹¹ * t) V/m
  3. Ey = -30*sqrt(2) sin(0.5 x 10¹¹ * x + 1.5 x 10³ * t) V/m
  4. Ey = -60 sin(0.5 x 10³ * x + 1.5 x 10¹¹ * t) V/m

Answer: Ey = -60 sin(0.5 x 10³ * x + 1.5 x 10¹¹ * t) V/m

E0 = c * B0 = 3x10⁸ * 2x10⁻⁷ = 60 V/m. The wave travels in -x direction. B is along z, propagation along -x: E x B must be along -x. With B along z-hat, E must be along -y-hat: Ey = -60 sin(...) V/m.

Q28. For an electromagnetic wave traveling through space, the electric field is given by E = E0 * sin(ky) * sin(omega*t) (in the z-direction). Find the corresponding magnetic field B in space. [c = speed of light in vacuum]

  1. (E0/c) * sin(ky) * sin(omega*t) in i-hat direction
  2. -(E0/c) * sin(ky) * sin(omega*t) in i-hat direction
  3. (E0/c) * cos(ky) * cos(omega*t) in i-hat direction
  4. -(E0/c) * cos(ky) * cos(omega*t) in i-hat direction

Answer: (E0/c) * cos(ky) * cos(omega*t) in i-hat direction

Using Faraday's law, (curl E)ₓ = dEz/dy = E0*k*cos(ky)*sin(omega*t) = -dBx/dt. Integrating: Bx = (E0/c)*cos(ky)*cos(omega*t), so B = (E0/c)*cos(ky)*cos(omega*t) i-hat.

Q29. A plane electromagnetic wave travels in the z-direction. Its electric field vector is given by: E(z,t) = E0*cos(kz - wt) x-hat + E0*cos(kz - wt + phi) y-hat, where E0 is the electric field amplitude, k is the wave number, w is the angular frequency, and phi is the phase difference between the x and y components. Which of the following statements are correct?

  1. If phi = pi/2, the wave is circularly polarized.
  2. If phi = pi, the wave is linearly polarized in the xy-plane.
  3. The magnetic field vector of the wave is perpendicular to both the electric field vector and the direction of propagation.
  4. For phi = pi/2, doubling the amplitude of the y-component of the electric field produces elliptical polarization.

Answer: The magnetic field vector of the wave is perpendicular to both the electric field vector and the direction of propagation.

With phi=pi/2 and equal amplitudes E0, the tip of E traces a circle, confirming circular polarization (A correct). With phi=pi, E = E0*cos(kz-wt)*(x-hat - y-hat), a linearly polarized wave along (x-y) direction (B correct). Option C is always true for any EM wave: B is perpendicular to E and to the propagation direction (C correct). For phi=pi/2 with y-amplitude doubled to 2E0, Ex=E0*cos and Ey=2E0*sin, giving an ellipse (D correct). All four options are correct; this is a multiple-correct type question.

Q30. In an RC charging circuit (capacitor initially uncharged), find the displacement current through the capacitor plates after a time equal to one time constant (tau = RC). Given the initial (maximum) current is 120 mA.

  1. 120/e mA
  2. 120/e² mA
  3. 120 mA
  4. None of these

Answer: 120/e mA

For a charging capacitor, the conduction current i = (V/R)*e^(-t/tau). Displacement current through the capacitor equals this conduction current (Maxwell's concept). At t = tau: i = (V/R)*e^(-1) = 120/e mA.

Q31. A plane electromagnetic wave travels in a medium with relative permeability mu_r = 1.6 and relative permittivity epsilon_r = 6.44. At a certain point the magnitude of the magnetic field intensity is B = 4.5 x 10⁻² T. Find the approximate magnitude of the electric field intensity at that point. (Speed of light in vacuum c = 3 x 10⁸ m/s)

  1. 16.9 V/m
  2. 2.25 x 10⁻² V/m
  3. 8.48 V/m
  4. 4.2 x 10⁶ V/m

Answer: 4.2 x 10⁶ V/m

The wave speed in the medium is v = c/sqrt(mu_r*epsilon_r) = 3e8/sqrt(10.304) ~ 9.35e7 m/s. Then E = v*B = 9.35e7 * 4.5e-2 ~ 4.2e6 V/m.

Q32. A material has conductivity sigma = 5.0 S/m and relative permittivity epsilon_r = 1. An electric field E = 250*sin(10¹⁰ * t) V/m exists in it. Identify the correct expressions for conduction current density Jc, displacement current density JD, and the frequency at which they are equal.

  1. Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz
  2. Jc = 22.1 cos(10¹⁰ t) A/m²; JD = 1250 sin(10¹⁰ t) A/m²; f = 90 GHz
  3. Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 sin(10¹⁰ t) A/m²; f = 50 GHz
  4. Jc = 1250 cos(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 50 GHz

Answer: Jc = 1250 sin(10¹⁰ t) A/m²; JD = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz

Jc = sigma*E = 5*250*sin(10¹⁰ t) = 1250*sin(10¹⁰ t). JD = epsilon₀*(dE/dt) = 8.85e-12 * 250*10¹⁰ * cos(10¹⁰ t) = 22.1*cos(10¹⁰ t). Amplitudes are equal when sigma = omega*epsilon, giving omega = sigma/epsilon₀ = 5/(8.85e-12) ≈ 5.65e11 rad/s => f ≈ 90 GHz.

Q33. An electromagnetic wave of frequency 1 x 10¹⁴ Hz propagates along the z-axis. The amplitude of the electric field is 4 V/m. Given epsilon₀ = 8.8 x 10⁻¹² C²/(N*m²), find the average energy density of the electric field.

  1. 35.2 x 10⁻¹⁰ J/m³
  2. 35.2 x 10⁻¹¹ J/m³
  3. 35.2 x 10⁻¹² J/m³
  4. 35.2 x 10⁻¹³ J/m³

Answer: 35.2 x 10⁻¹² J/m³

The average energy density of the electric field in an EM wave is u_E = (1/2)*epsilon₀*E_rms² = (1/2)*epsilon₀*(E0²/2) = epsilon₀*E0²/4. Substituting: (8.8e-12 * 16)/4 = 35.2e-12 J/m³.

Q34. An electromagnetic wave has electric field E = 4.8*10² * cos(2*10⁷ * z + 6*10¹⁵ * t) * (-i_hat + 2*j_hat) V/m. Find the associated magnetic field B.

  1. B = 1.6*10^(-6) * cos(2*10⁷*z + 6*10¹⁵*t) * (2*i_hat - j_hat) Wb/m²
  2. B = 1.6*10^(-6) * cos(2*10⁷*z + 6*10¹⁵*t) * (-i_hat + 2*j_hat) Wb/m²
  3. B = 1.6*10^(-6) * cos(2*10⁷*z + 6*10¹⁵*t) * (2*i_hat + j_hat) Wb/m²
  4. B = 1.6*10^(-6) * cos(2*10⁷*z + 6*10¹⁵*t) * (2*i_hat - j_hat) Wb/m²

Answer: B = 1.6*10^(-6) * cos(2*10⁷*z + 6*10¹⁵*t) * (2*i_hat + j_hat) Wb/m²

The wave travels in the -z direction (positive kz with positive omega*t indicates -z propagation). B = (k_hat x E)/c = (-z_hat x E)/c. Computing: -z_hat x (-i+2j) = z_hat x (i-2j) = z_hat x i - 2*(z_hat x j) = j_hat - 2*(-i_hat) = 2i+j. Magnitude: |B| = |E|/c = 4.8*10² * sqrt(5) / (3*10⁸ * sqrt(5)) = 1.6*10⁻⁶. So B = 1.6*10⁻⁶*(2i+j)*cos(...).

Q35. In a material with conductivity sigma = 5.0 S/m and relative permittivity epsilon_r = 1, the electric field is E = 250 sin(10¹⁰ * t) V/m. Determine the conduction current density Jc, the displacement current density Jd, and the frequency f at which their amplitudes are equal.

  1. Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz
  2. Jc = 22.1 cos(10¹⁰ t) A/m²; Jd = 1250 sin(10¹⁰ t) A/m²; f = 90 GHz
  3. Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 sin(10¹⁰ t) A/m²; f = 50 GHz
  4. Jc = 1250 cos(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 50 GHz

Answer: Jc = 1250 sin(10¹⁰ t) A/m²; Jd = 22.1 cos(10¹⁰ t) A/m²; f = 90 GHz

Jc = 5 * 250 sin(10¹⁰ t) = 1250 sin(10¹⁰ t) A/m². Jd = epsilon₀ * dE/dt = 8.854e-12 * 250 * 10¹⁰ * cos(10¹⁰ t) = 22.1 cos(10¹⁰ t) A/m². For equal amplitudes: sigma = omega*epsilon₀ -> f = sigma/(2*pi*epsilon₀) = 5/(2*pi*8.854e-12) approx 90 GHz.

Q36. A linearly polarized electromagnetic wave E_vec = E0 * i_hat * cos(k*z - omega*t) is incident normally on a perfectly reflecting infinite wall at z = 0. Assuming the wall is optically inactive, what is the expression for the reflected wave?

  1. E_r_vec = -E0 * i_hat * cos(k*z - omega*t)
  2. E_r_vec = E0 * i_hat * cos(k*z + omega*t)
  3. E_r_vec = -E0 * i_hat * cos(k*z + omega*t)
  4. E_r_vec = E0 * i_hat * sin(k*z - omega*t)

Answer: E_r_vec = -E0 * i_hat * cos(k*z + omega*t)

On reflection from a perfectly conducting surface, the electric field undergoes a phase shift of 180 deg (sign reversal). The reflected wave travels in the +z direction reversed to -z... wait: incident travels in +z, reflected travels in -z. The reflected wave: E_r = -E0*i_hat*cos(k*z + omega*t) (propagating in -z direction with pi phase shift).

Q37. The electric field intensity produced by radiation from a 100 W bulb at a distance of 3 m is E. What is the electric field intensity produced by a 50 W bulb at the same distance?

  1. E/2
  2. 2E
  3. E/sqrt(2)
  4. sqrt(2)*E

Answer: E/sqrt(2)

Electric field amplitude E is proportional to sqrt(intensity) = sqrt(P/(4*pi*r²)) ~ sqrt(P) at fixed distance. So E₅₀/E₁₀₀ = sqrt(50/100) = 1/sqrt(2). Hence E₅₀ = E/sqrt(2).

Q38. An electromagnetic wave consists of periodically oscillating electric and magnetic field vectors. How are these vectors oriented relative to each other and what is their phase relationship?

  1. in randomly oriented planes but vibrating in phase.
  2. in mutually perpendicular planes but vibrating in phase.
  3. in mutually perpendicular planes but vibrating with a phase difference of pi.
  4. in mutually perpendicular planes but vibrating with a phase difference of pi/2.

Answer: in mutually perpendicular planes but vibrating in phase.

In a plane electromagnetic wave, the electric field E and magnetic field B are mutually perpendicular to each other and to the direction of propagation. From Maxwell's equations, both fields vary as sin(kx - omega*t) — they are always in phase (phase difference = 0).

Q39. A floodlight filtered to emit only red light produces a sinusoidal plane electromagnetic wave with electric field Eₓ = 36*sin(1.20*10⁷ * z - 3.6*10¹⁵ * t) V/m. What is the average intensity of this beam?

  1. 0.86 W/m²
  2. 1.72 W/m²
  3. 3.44 W/m²
  4. 6.88 W/m²

Answer: 1.72 W/m²

The amplitude of the electric field is E₀ = 36 V/m. Average intensity = E₀² / (2*Z₀) where Z₀ = mu₀*c = 377 ohm. I = 36² / (2*377) = 1296/754 ≈ 1.72 W/m².

Q40. An electromagnetic wave of frequency 1 x 10¹⁴ Hz propagates along the z-axis with an electric field amplitude of 4 V/m. Given epsilon₀ = 8.8 x 10⁻¹² C²/(N*m²), what is the average energy density of the electric field?

  1. 35.2 × 10⁻¹⁰ J/m³
  2. 35.2 × 10⁻¹¹ J/m³
  3. 35.2 × 10⁻¹² J/m³
  4. 35.2 × 10⁻¹³ J/m³

Answer: 35.2 × 10⁻¹² J/m³

For a sinusoidal EM wave, the time-averaged electric energy density is u_E = (1/4)*epsilon₀*E0² = (1/4) * 8.8e-12 * 16 = 35.2e-12 J/m³.

Q41. A plane electromagnetic wave of angular frequency omega propagates in a poorly conducting medium with conductivity sigma and relative permittivity epsilon. Find the ratio of conduction current density to displacement current density in the medium.

  1. epsilon*epsilon₀*omega/sigma
  2. sigma/(epsilon*epsilon₀*omega)
  3. omega/(sigma*epsilon₀*epsilon)
  4. omega*sigma/(epsilon₀*epsilon)

Answer: sigma/(epsilon*epsilon₀*omega)

For a sinusoidal EM wave E = E₀*cos(omega*t): J_c = sigma*E, J_d = epsilon*epsilon₀*(dE/dt) = epsilon*epsilon₀*omega*E₀. The ratio |J_c|/|J_d| = sigma/(epsilon*epsilon₀*omega).

Q42. The electric field of a plane polarised electromagnetic wave in free space at t = 0 is given by E(x, z) = 10 * j_hat * cos(6x + 8z). Find the magnetic field B(x, z, t), where c is the speed of light.

  1. (1/c) * (6*k_hat + 8*i_hat) * cos[(6x - 8z + 10ct)]
  2. (1/c) * (6*k_hat - 8*i_hat) * cos[(6x + 8z - 10ct)]
  3. (1/c) * (6*k_hat + 8*i_hat) * cos[(6x + 8z - 10ct)]
  4. (1/c) * (6*k_hat - 8*i_hat) * cos[(6x + 8z + 10ct)]

Answer: (1/c) * (6*k_hat - 8*i_hat) * cos[(6x + 8z - 10ct)]

Wave vector k_vec = 6i + 8k, |k_vec| = 10, so omega = 10c. Unit vector k_hat = (6i+8k)/10. B_hat = k_hat x E_hat = ((6i+8k)/10) x j = (6(i x j) + 8(k x j))/10 = (6k - 8i)/10. Magnitude |B| = |E|/c = 10/c. So B = (10/c)*(6k-8i)/10 * cos(6x+8z-10ct) = (1/c)*(6k-8i)*cos(6x+8z-10ct).

Q43. An observer stands 90 cm from a point source of light with power 250 W. What is the rms value of the electric field at the observer's position?

  1. 48 V/m
  2. 96 V/m
  3. 40 V/m
  4. 80 V/m

Answer: 96 V/m

Intensity I = 250/(4*pi*0.81) ≈ 24.56 W/m². Using I = epsilon0*c*E_rms²: E_rms = sqrt(24.56/(8.85e-12 * 3e8)) = sqrt(24.56/2.655e-3) ≈ sqrt(9250) ≈ 96 V/m.

Q44. An electromagnetic wave travels along the z-axis: E = E₀ cos(k*z - omega*t). Which of the following options are correct?

  1. The associated magnetic field is given as B = (1/c) k_hat x E = (1/omega)(k_vec x E)
  2. The electromagnetic field can be written in terms of the associated magnetic field as E = c*(B x k_hat)
  3. k_hat dot E = 0, k_hat dot B = 0
  4. k_hat x E = 0, k_hat x B = 0

Answer: The associated magnetic field is given as B = (1/c) k_hat x E = (1/omega)(k_vec x E)

For an EM wave propagating in k_hat direction: B = (k_hat x E)/c = (k_vec x E)/omega (since k_vec = k*k_hat and omega = k*c). Also E = c*(B x k_hat). Since E is transverse, k_hat dot E = 0 and k_hat dot B = 0. But k_hat x E != 0. So options 1, 2, 3 are correct; option 4 is wrong.

Q45. A plane electromagnetic wave propagates along the x-direction. Which of the following pairs of E and B field components is/are physically consistent with this propagation?

  1. Eₓ, B_y.
  2. E_y, B_z.
  3. Bₓ, E_y.
  4. E_z, B_y.

Answer: E_y, B_z.

Options B (E_y, B_z) and D (E_z, B_y) are consistent with propagation along the x-axis. Options A and C involve longitudinal field components, which are not allowed in EM waves.

Q46. A parallel plate capacitor with circular plates of radius R is being charged by a constant current I. Which of the following statements about the fields and energy are correct? (A) The Poynting vector points radially inward everywhere on a cylindrical surface at the rim of the plates. (B) The magnetic field between the plates is clockwise when viewed from above (current flowing upward). (C) The magnetic energy density between the plates is increasing with time. (D) The net Poynting vector flux into the capacitor equals the rate of change of stored energy.

  1. (A) Poynting vector points radially inward everywhere on the circle.
  2. (B) Magnetic field is clockwise as seen from above.
  3. (C) Magnetic energy density is increasing with time.
  4. (D) Poynting vector flux is equal to the rate of change of stored energy.

Answer: (D) Poynting vector flux is equal to the rate of change of stored energy.

The magnetic field due to displacement current (counterclockwise from above for upward current) combined with upward E gives S = E x B pointing radially inward. Poynting theorem guarantees flux of S = rate of energy storage. B is false (field is counterclockwise, not clockwise).

Q47. Charges in an antenna undergo oscillations with amplitude x0 and angular frequency omega. The average power P radiated depends on charge q, x0, omega, epsilon0, and the speed of light c. Using dimensional analysis, the correct expression for P (where a is a dimensionless constant) is:

  1. a*q²*omega⁴*x0² / (epsilon0*c²)
  2. a*q²*omega⁴*x0² / (epsilon0*c³)
  3. a*q⁴*omega²*x0 / (epsilon0²*c⁶)
  4. a*q⁴*omega⁴*x0² / (epsilon0²*c⁴)

Answer: a*q²*omega⁴*x0² / (epsilon0*c³)

Setting the dimensions of option B equal to ML²T⁻³ confirms it is the only dimensionally consistent formula for radiated power.

Q48. An electromagnetic wave travels along the z-axis: E = E0*cos(kz - omega*t). Which of the following are correct? (A) B = (1/c)*(k_hat cross E) = (1/omega)*(k_vec cross E) (B) E = c*(B cross k_hat) (C) k_hat dot E = 0, k_hat dot B = 0 (D) k_hat cross E = 0, k_hat cross B = 0

  1. B = (1/c)*(k_hat cross E) = (1/omega)*(k_vec cross E)
  2. E = c*(B cross k_hat)
  3. k_hat dot E = 0, k_hat dot B = 0
  4. k_hat cross E = 0, k_hat cross B = 0

Answer: B = (1/c)*(k_hat cross E) = (1/omega)*(k_vec cross E)

For a plane EM wave along z: (A) B = (1/c)(k_hat cross E); since k_vec = k*k_hat and k/omega = 1/c, also B = (1/omega)(k_vec cross E). Correct. (B) From B = (1/c)(k_hat cross E), using vector identity: E = c(B cross k_hat). Correct. (C) k_hat perpendicular to both E and B (transverse wave) => dot products are zero. Correct. (D) k_hat cross E = 0 and k_hat cross B = 0 would mean E, B parallel to k_hat — INCORRECT for transverse EM waves. Correct options: A, B, C.

Q49. A plane electromagnetic wave propagates along the x-direction. Which of the following pairs of E and B field components are POSSIBLE?

  1. Eₓ, B_y
  2. E_y, B_z
  3. Bₓ, E_y
  4. E_z, B_y

Answer: E_y, B_z

E and B must both be transverse for an EM wave. Options with Eₓ or Bₓ are invalid. E_y with B_z gives E x B along +x, consistent with propagation in +x. E_z with B_y gives propagation in -x. For propagation along +x direction: E_y, B_z is the correct pair.

Q50. A capacitor with circular plates of radius R is charged by a constant current I. Which of the following statements are correct? (A) The Poynting vector points radially inward everywhere on the cylindrical surface around the capacitor. (B) The magnetic field inside the capacitor is clockwise as viewed from above (in the direction of current flow). (C) The magnetic energy density inside the capacitor is increasing with time. (D) The flux of the Poynting vector through the cylindrical surface equals the rate of change of stored electric energy.

  1. Poynting vector points radially inward everywhere on the circle.
  2. Magnetic field is clockwise as seen from above.
  3. Magnetic energy density is increasing with time.
  4. Poynting vector flux is equal to the rate of change of stored energy.

Answer: Poynting vector flux is equal to the rate of change of stored energy.

Poynting vector S = (1/mu0)(E x B) points radially inward (energy flows into the capacitor). Magnetic field from constant displacement current I is constant, so magnetic energy density is constant, not increasing. Poynting flux equals rate of change of stored (electric) energy. Options A and D are correct.

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