Category: HC Verma Part 2: Modern Physics
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HC Verma Chapter 40 Problem 10 — Intensity from Electric Field Amplitude
Problem Statement Determine the electric field for the configuration described: In an electromagnetic wave travelling through free space, the electric field amplitude is $E_0 = 120$ V/m. Calculate the intensity of the wave. $I = \frac{E_0^2}{2\mu_0 c}$; impedance of free space $Z_0 = \mu_0 c = 377\;\Omega$ Step 1: $$I = \frac{E_0^2}{2\mu_0 c} = \frac{(120)^2}{2\times377}…
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HC Verma Chapter 40 Problem 9 — Wavelength Range of X-rays
Problem Statement Solve the oscillation/wave problem: X-rays have frequencies ranging from $3\times10^{16}$ Hz to $3\times10^{19}$ Hz. Find the corresponding range of wavelengths. $\lambda = c/f$ Step 1: For $f_{min} = 3\times10^{16}$ Hz (longest wavelength): $$\lambda_{max} = \frac{3\times10^8}{3\times10^{16}} = 10^{-8}\text{ m} = 10\text{ nm}$$ Step Given Information $\lambda_{max} = \frac{3\times10^8}{3\times10^{16}} = 10^{-8}\text{ m} = 10\text{ nm}$…
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HC Verma Chapter 40 Problem 8 — Displacement Current Between Capacitor Plates
Problem Statement Solve the capacitor/capacitance problem: A parallel-plate capacitor with plate area $A = 0.02$ m² has an electric field between the plates increasing at $dE/dt = 10^{12}$ V/(m·s). Find the displacement current flowing between the plates. Displacement current: $I_d = \varepsilon_0\frac{d\Phi_E}{dt} = \varepsilon_0 A\frac{dE}{dt}$ Step 1: $$ Given Information See problem statement for all…
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HC Verma Chapter 40 Problem 7 — Radiation Pressure on Absorbing Surface
Problem Statement Solve the fluid mechanics problem: Sunlight of intensity $I = 1400$ W/m² falls on a perfectly absorbing surface. Calculate the radiation pressure exerted on the surface. For perfect absorption, radiation pressure $P = I/c$ For perfect reflection: $P = 2I/c$ Step 1: Since surface is perfectly absorbing: $$P = \frac{I}{c} = \frac{1400} Given…
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HC Verma Chapter 40 Problem 6 — Electric Field Amplitude of Sunlight
Problem Statement Determine the electric field for the configuration described: The intensity of sunlight reaching Earth’s surface is $I = 1400$ W/m². Find the amplitude of the electric field $E_0$ in this radiation. $I = \frac{1}{2}\varepsilon_0 c E_0^2 \Rightarrow E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}}$ Step 1: $$E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2\times1 Given Information See problem…
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HC Verma Chapter 40 Problem 5 — Energy Density of EM Wave
Problem Statement Solve the oscillation/wave problem: Find the average energy density of an electromagnetic wave whose electric field amplitude is $E_0 = 100$ V/m. Average electric energy density: $u_E = \frac{1}{2}\varepsilon_0 E_0^2/2 = \frac{\varepsilon_0 E_0^2}{4}$ wait, more carefully: $u_E = \frac{1}{2}\varepsilon_0 E_{rms}^2 = \frac{1}{2}\vareps Given Information See problem statement for all given quantities. Physical Concepts…
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HC Verma Chapter 40 Problem 4 — Magnetic Field Amplitude in EM Wave
Problem Statement Solve the magnetic field/force problem: The electric field amplitude in an electromagnetic wave travelling in free space is $E_0 = 50$ N/C. Find the amplitude of the magnetic field $B_0$. In a plane EM wave: $B_0 = E_0/c$ Step 1: The ratio of electric to magnetic amplitudes equals $c$: $$B_0 = \frac{E_0}{c} =…
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HC Verma Chapter 40 Problem 3 — Frequency of Sodium Yellow Light
Problem Statement Calculate the frequency of sodium yellow light with wavelength $\lambda = 589$ nm. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful…
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HC Verma Chapter 40 Problem 2 — Wavelength of AM Radio Wave
Problem Statement Solve the oscillation/wave problem: An AM radio station broadcasts at a frequency of 1000 kHz. Find the wavelength of the radio wave. $c = f\lambda \Rightarrow \lambda = c/f$ Step 1: $f = 1000\text{ kHz} = 10^6\text{ Hz}$ Step 2: $$\lambda = \frac{c}{f} = \frac{3\times10^8\text{ m/s}}{10^6\text{ Hz}} = 300\text{ m}$$ $$\boxed{\lambda Given Information…
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HC Verma Chapter 40 Problem 1 — Speed of Electromagnetic Wave
Problem Statement Solve the magnetic field/force problem: Show that the speed of an electromagnetic wave in vacuum is $c = 1/\sqrt{\mu_0 \varepsilon_0}$ and calculate its value given $\mu_0 = 4\pi\times10^{-7}$ T·m/A and $\varepsilon_0 = 8.85\times10^{-12}$ C²/(N·m²). Maxwell’s wave equations give $c = 1/\sqrt{\mu_0\varepsilon_0}$ Step 1: Product $\mu_0\va Given Information $c = 1/$ $c = 1/$…