How to Use This Problem Set
Your full workout for Electromagnetic Waves, grouped by theme: displacement current, plane-wave anatomy (E0/B0, k, omega, directions), speed in media, and spectrum conversions/identification.
Keep these handy:
i d = ε 0 d Φ E d t = ε 0 A d E d t = C d V d t i_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = \varepsilon_0 A\dfrac{dE}{dt} = C\dfrac{dV}{dt} i d = ε 0 d t d Φ E = ε 0 A d t d E = C d t d V ∮ B ⃗ ⋅ d l ⃗ = μ 0 ( i c + i d ) \oint\vec B\cdot d\vec l = \mu_0(i_c + i_d) ∮ B ⋅ d l = μ 0 ( i c + i d ) Inside the gap (r < R): B = μ 0 i d r 2 π R 2 B = \dfrac{\mu_0 i_d r}{2\pi R^2} B = 2 π R 2 μ 0 i d r
E x = E 0 sin ( k z − ω t ) E_x = E_0\sin(kz-\omega t) E x = E 0 sin ( k z − ω t ) B y = B 0 sin ( k z − ω t ) B_y = B_0\sin(kz-\omega t) B y = B 0 sin ( k z − ω t ) k = 2 π / λ k = 2\pi/\lambda k = 2 π / λ ω = c k \omega = ck ω = c k ν λ = c \nu\lambda = c ν λ = c B 0 = E 0 / c B_0 = E_0/c B 0 = E 0 / c E ^ × B ^ \hat E\times\hat B E ^ × B ^ c = 1 / μ 0 ε 0 = 3 × 10 8 c = 1/\sqrt{\mu_0\varepsilon_0} = 3\times10^8 c = 1/ μ 0 ε 0 = 3 × 1 0 8 v = c / μ r ε r v = c/\sqrt{\mu_r\varepsilon_r} v = c / μ r ε r Spectrum (increasing frequency): radio, microwave, IR, visible (700-400 nm), UV, X, gamma
Carry units; compute cross products, never guess directions.
Solved Examples - Displacement Current
Example 1. A capacitor charges with 0.25 A in its leads. The displacement current in the gap?
Solution: i d = d Q / d t = 0.25 i_d = dQ/dt = 0.25 i d = d Q / d t = 0.25
Example 2. A 2 μ \mu μ 2 × 10 5 2\times10^5 2 × 1 0 5 i d i_d i d
Solution: i d = C d V / d t = 2 × 10 − 6 × 2 × 10 5 = 0.4 i_d = C\,dV/dt = 2\times10^{-6}\times2\times10^5 = 0.4 i d = C d V / d t = 2 × 1 0 − 6 × 2 × 1 0 5 = 0.4
Example 3. E between plates of area 0.02 0.02 0.02 2 ^2 2 5 × 10 11 5\times10^{11} 5 × 1 0 11 i d i_d i d
Solution: i d = ε 0 A d E / d t = 8.85 × 10 − 12 × 0.02 × 5 × 10 11 = 8.85 × 10 − 2 i_d = \varepsilon_0 A\,dE/dt = 8.85\times10^{-12}\times0.02\times5\times10^{11} = 8.85\times10^{-2} i d = ε 0 A d E / d t = 8.85 × 1 0 − 12 × 0.02 × 5 × 1 0 11 = 8.85 × 1 0 − 2
Example 4. 100 pF on V = 230 sin ( 300 t ) V = 230\sin(300t) V = 230 sin ( 300 t ) i d i_d i d
Solution: i d m a x = C V 0 ω = 10 − 10 × 230 × 300 = 6.9 μ i_d^{max} = CV_0\omega = 10^{-10}\times230\times300 = 6.9\ \mu i d ma x = C V 0 ω = 1 0 − 10 × 230 × 300 = 6.9 μ
Example 5. For plates of radius R = 6 cm carrying i d i_d i d μ \mu μ
Solution: B = μ 0 i d r 2 π R 2 = 2 × 10 − 7 × 6.9 × 10 − 6 × 0.03 3.6 × 10 − 3 ≈ 1.15 × 10 − 11 B = \dfrac{\mu_0 i_d r}{2\pi R^2} = \dfrac{2\times10^{-7}\times6.9\times10^{-6}\times0.03}{3.6\times10^{-3}} \approx 1.15\times10^{-11} B = 2 π R 2 μ 0 i d r = 3.6 × 1 0 − 3 2 × 1 0 − 7 × 6.9 × 1 0 − 6 × 0.03 ≈ 1.15 × 1 0 − 11
Example 6. Where is i d i_d i d
Solution: In the gap (i d = i i_d = i i d = i i c = 0 i_c = 0 i c = 0 i d = 0 i_d = 0 i d = 0 i c = i i_c = i i c = i
Solved Examples - Plane-Wave Anatomy
Example 7. (NCERT 8.1) E = 6.3 j ^ \hat j j ^ B ⃗ \vec B B
Solution: B = E / c = 2.1 × 10 − 8 B = E/c = 2.1\times10^{-8} B = E / c = 2.1 × 1 0 − 8 j ^ × k ^ = i ^ \hat j\times\hat k = \hat i j ^ × k ^ = i ^ B ⃗ = 2.1 × 10 − 8 k ^ \vec B = 2.1\times10^{-8}\hat k B = 2.1 × 1 0 − 8 k ^
Example 8. E 0 = 90 E_0 = 90 E 0 = 90 B 0 B_0 B 0
Solution: B 0 = E 0 / c = 90 3 × 10 8 = 3 × 10 − 7 B_0 = E_0/c = \frac{90}{3\times10^8} = 3\times10^{-7} B 0 = E 0 / c = 3 × 1 0 8 90 = 3 × 1 0 − 7
Example 9. B 0 = 5 × 10 − 7 B_0 = 5\times10^{-7} B 0 = 5 × 1 0 − 7 E 0 E_0 E 0
Solution: E 0 = c B 0 = 150 E_0 = cB_0 = 150 E 0 = c B 0 = 150
Example 10. (NCERT 8.2) B y = 2 × 10 − 7 sin ( 0.5 × 10 3 x + 1.5 × 10 11 t ) B_y = 2\times10^{-7}\sin(0.5\times10^3x+1.5\times10^{11}t) B y = 2 × 1 0 − 7 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) λ \lambda λ ν \nu ν
Solution: λ = 2 π / 500 = 1.26 \lambda = 2\pi/500 = 1.26 λ = 2 π /500 = 1.26 ν = ω / 2 π = 23.9 \nu = \omega/2\pi = 23.9 ν = ω /2 π = 23.9 E z = 60 sin ( 0.5 × 10 3 x + 1.5 × 10 11 t ) E_z = 60\sin(0.5\times10^3x+1.5\times10^{11}t) E z = 60 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) − x -x − x
Example 11. A wave of frequency 25 MHz: find λ \lambda λ ω \omega ω
Solution: λ = 12 \lambda = 12 λ = 12 k = 2 π / 12 = 0.52 k = 2\pi/12 = 0.52 k = 2 π /12 = 0.52 ω = 2 π ν = 1.57 × 10 8 \omega = 2\pi\nu = 1.57\times10^8 ω = 2 π ν = 1.57 × 1 0 8
Example 12. E along i ^ \hat i i ^ j ^ \hat j j ^
Solution: i ^ × j ^ = k ^ \hat i\times\hat j = \hat k i ^ × j ^ = k ^ +z .
Example 13. E along k ^ \hat k k ^ j ^ \hat j j ^
Solution: Need k ^ × B ^ = j ^ \hat k\times\hat B = \hat j k ^ × B ^ = j ^ B ^ = i ^ \hat B = \hat i B ^ = i ^ k ^ × i ^ = j ^ \hat k\times\hat i = \hat j k ^ × i ^ = j ^
Example 14. When E at a point is zero (in a travelling wave), B is:
Solution: Zero too — E and B oscillate in phase , sharing sin ( k z − ω t ) \sin(kz-\omega t) sin ( k z − ω t )
Example 15. Compute c from μ 0 = 4 π × 10 − 7 \mu_0 = 4\pi\times10^{-7} μ 0 = 4 π × 1 0 − 7 ε 0 = 8.85 × 10 − 12 \varepsilon_0 = 8.85\times10^{-12} ε 0 = 8.85 × 1 0 − 12
Solution: μ 0 ε 0 = 1.11 × 10 − 17 \mu_0\varepsilon_0 = 1.11\times10^{-17} μ 0 ε 0 = 1.11 × 1 0 − 17 c = 1 / 1.11 × 10 − 17 ≈ 3.0 × 10 8 c = 1/\sqrt{1.11\times10^{-17}} \approx 3.0\times10^8 c = 1/ 1.11 × 1 0 − 17 ≈ 3.0 × 1 0 8
Example 16. Light in glass (n = 1.5): speed?
Solution: v = c / n = 2 × 10 8 v = c/n = 2\times10^8 v = c / n = 2 × 1 0 8
Example 17. A medium with μ r = 1 \mu_r = 1 μ r = 1 c / 2 c/2 c /2 ε r \varepsilon_r ε r
Solution: n = μ r ε r = 2 n = \sqrt{\mu_r\varepsilon_r} = 2 n = μ r ε r = 2 ε r = 4 \varepsilon_r = 4 ε r = 4
Example 18. Does the frequency change when light enters glass?
Solution: No — frequency is set by the source; the wavelength shrinks (λ = v / ν \lambda = v/\nu λ = v / ν
Solved Examples - Spectrum Conversions
Example 19. AM station at 1000 kHz: wavelength?
Solution: λ = 3 × 10 8 / 10 6 = 300 \lambda = 3\times10^8/10^6 = 300 λ = 3 × 1 0 8 /1 0 6 = 300
Example 20. Microwave oven wavelength 12.2 cm: frequency?
Solution: ν = 3 × 10 8 / 0.122 ≈ 2.5 \nu = 3\times10^8/0.122 \approx 2.5 ν = 3 × 1 0 8 /0.122 ≈ 2.5
Example 21. 0.1 nm X-ray: frequency?
Solution: ν = 3 × 10 8 / 10 − 10 = 3 × 10 18 \nu = 3\times10^8/10^{-10} = 3\times10^{18} ν = 3 × 1 0 8 /1 0 − 10 = 3 × 1 0 18
Example 22. 500 nm light: frequency?
Solution: ν = 3 × 10 8 / 5 × 10 − 7 = 6 × 10 14 \nu = 3\times10^8/5\times10^{-7} = 6\times10^{14} ν = 3 × 1 0 8 /5 × 1 0 − 7 = 6 × 1 0 14
Example 23. A 3 m wave and a 3 cm wave: name the bands.
Solution: 3 m → radio (FM region); 3 cm → microwave (radar region).
Example 24. Order by frequency: 600 m, 10 μ \mu μ
Solution: 5 × 10 5 5\times10^5 5 × 1 0 5 3 × 10 13 3\times10^{13} 3 × 1 0 13 6 × 10 14 6\times10^{14} 6 × 1 0 14 6 × 10 18 6\times10^{18} 6 × 1 0 18
Example 25. Identify by production: (a) magnetron, (b) radioactive cobalt nucleus, (c) hot filament's molecular vibration, (d) FM aerial.
Solution: (a) microwave; (b) gamma; (c) infrared; (d) radio.
Example 26. Identify by detector: thermopile; Geiger tube; the eye; receiver aerial.
Solution: IR; X/gamma; visible; radio — Table 8.1 rows.
Example 27. Which band: tanning, ozone-absorbed, blocked by window glass?
Solution: Ultraviolet on all three counts.
Solved Examples - Mixed JEE/NEET Patterns
Example 28. A wave E = 60 sin ( k z − 6 × 10 15 t ) E = 60\sin(kz - 6\times10^{15}t) E = 60 sin ( k z − 6 × 1 0 15 t ) ν \nu ν λ \lambda λ
Solution: ν = ω / 2 π = 9.5 × 10 14 \nu = \omega/2\pi = 9.5\times10^{14} ν = ω /2 π = 9.5 × 1 0 14 λ = c / ν ≈ 314 \lambda = c/\nu \approx 314 λ = c / ν ≈ 314 ultraviolet (just past violet).
Example 29. B 0 B_0 B 0
Solution: B 0 = 60 3 × 10 8 = 2 × 10 − 7 B_0 = \frac{60}{3\times10^8} = 2\times10^{-7} B 0 = 3 × 1 0 8 60 = 2 × 1 0 − 7
Example 30. A 90 MHz wave enters a medium with ε r = 2.25 \varepsilon_r = 2.25 ε r = 2.25 μ r = 1 \mu_r = 1 μ r = 1
Solution: v = c / 1.5 = 2 × 10 8 v = c/1.5 = 2\times10^8 v = c /1.5 = 2 × 1 0 8 λ = v / ν = 2 × 10 8 / 9 × 10 7 = 2.2 \lambda = v/\nu = 2\times10^8/9\times10^7 = 2.2 λ = v / ν = 2 × 1 0 8 /9 × 1 0 7 = 2.2
Example 31. Time for a radar pulse to return from an aircraft 30 km away?
Solution: t = 2 d / c = 2 × 3 × 10 4 3 × 10 8 = 2 × 10 − 4 t = 2d/c = \dfrac{2\times3\times10^4}{3\times10^8} = 2\times10^{-4} t = 2 d / c = 3 × 1 0 8 2 × 3 × 1 0 4 = 2 × 1 0 − 4
Example 32. Two waves: 2 MHz and 2 GHz. Which diffracts around hills better, and why?
Solution: The 2 MHz wave (λ = 150 \lambda = 150 λ = 150 λ \lambda λ λ = 15 \lambda = 15 λ = 15
Example 33. The earth receives 1.5 ms-old radio echoes from a satellite. Its distance?
Solution: d = c t 2 = 3 × 10 8 × 1.5 × 10 − 3 2 = 2.25 × 10 5 d = \dfrac{ct}{2} = \dfrac{3\times10^8\times1.5\times10^{-3}}{2} = 2.25\times10^5 d = 2 c t = 2 3 × 1 0 8 × 1.5 × 1 0 − 3 = 2.25 × 1 0 5
Example 34. In vacuum, do gamma rays outrun radio waves?
Solution: No — all EM waves travel at exactly c in vacuum , independent of wavelength (so precisely that c defines the metre). They differ in frequency, not speed.
