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Problem 1: Electromagnetic Waves in a Medium

Question:An electromagnetic wave propagates through a non-conductive medium with a relative permittivity εr=4\varepsilon_r = 4εr​=4 and relative permeability μr=2\mu_r = 2μr​=2. The electric field of the wave is given by:

E(x,t)=10cos⁡(2π×109t−10πx) y^ (V/m).E(x, t) = 10 \cos(2\pi \times 10^9 t - 10\pi x) \, \hat{y} \, \text{(V/m)}.E(x,t)=10cos(2π×109t−10πx)y^​(V/m).

(a) Determine the speed of the wave in the medium.(b) Calculate the wavelength of the wave.(c) Find the corresponding magnetic field B(x,t)B(x, t)B(x,t).

Solution:(a) Speed of the WaveThe speed of an electromagnetic wave in a medium is given by:

v=cεrμr,v = \frac{c}{\sqrt{\varepsilon_r \mu_r}},v=εr​μr​​c​,

where ccc is the speed of light in a vacuum (c≈3×108 m/sc \approx 3 \times 10^8 \, \text{m/s}c≈3×108m/s). Substituting the given values:

v=3×1084×2=3×1088≈1.06×108 m/s.v = \frac{3 \times 10^8}{\sqrt{4 \times 2}} = \frac{3 \times 10^8}{\sqrt{8}} \approx 1.06 \times 10^8 \, \text{m/s}.v=4×2​3×108​=8​3×108​≈1.06×108m/s.

(b) WavelengthThe wavelength λ\lambdaλ of the wave is related to its speed vvv and frequency fff:

λ=vf.\lambda = \frac{v}{f}.λ=fv​.

The frequency fff is 109 Hz10^9 \, \text{Hz}109Hz. Thus:

λ=1.06×108109=0.106 m.\lambda = \frac{1.06 \times 10^8}{10^9} = 0.106 \, \text{m}.λ=1091.06×108​=0.106m.

(c) Magnetic FieldThe magnetic field B(x,t)B(x, t)B(x,t) is related to the electric field E(x,t)E(x, t)E(x,t) by:

B(x,t)=E(x,t)v z^.B(x, t) = \frac{E(x, t)}{v} \, \hat{z}.B(x,t)=vE(x,t)​z^.

Given E(x,t)=10cos⁡(2π×109t−10πx)E(x, t) = 10 \cos(2\pi \times 10^9 t - 10\pi x)E(x,t)=10cos(2π×109t−10πx), we find:

B(x,t)=101.06×108cos⁡(2π×109t−10πx) z^.B(x, t) = \frac{10}{1.06 \times 10^8} \cos(2\pi \times 10^9 t - 10\pi x) \, \hat{z}.B(x,t)=1.06×10810​cos(2π×109t−10πx)z^.

Simplifying:

B(x,t)≈9.43×10−8cos⁡(2π×109t−10πx) z^ (T).B(x, t) \approx 9.43 \times 10^{-8} \cos(2\pi \times 10^9 t - 10\pi x) \, \hat{z} \, \text{(T)}.B(x,t)≈9.43×10−8cos(2π×109t−10πx)z^(T).

Problem 2: Quantum Mechanics – Infinite Square Well

Question:A particle of mass mmm is confined in an infinite potential well of width LLL. The wave function of the particle in the ground state is given by:

ψ(x)=2Lsin⁡(πxL), 0≤x≤L.\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right), \, 0 \leq x \leq L.ψ(x)=L2​​sin(Lπx​),0≤x≤L.

(a) Calculate the probability of finding the particle in the first quarter of the well (0≤x≤L/40 \leq x \leq L/40≤x≤L/4).(b) Determine the energy of the particle in the ground state.

Solution:(a) Probability CalculationThe probability PPP of finding the particle in a region is given by:

P=∫ab∣ψ(x)∣2dx.P = \int_a^b |\psi(x)|^2 dx.P=∫ab​∣ψ(x)∣2dx.

For 0≤x≤L/40 \leq x \leq L/40≤x≤L/4:

P=∫0L/4(2Lsin⁡(πxL))2dx.P = \int_0^{L/4} \left(\sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)\right)^2 dx.P=∫0L/4​(L2​​sin(Lπx​))2dx.

Expanding the integral:

P=2L∫0L/4sin⁡2(πxL)dx.P = \frac{2}{L} \int_0^{L/4} \sin^2\left(\frac{\pi x}{L}\right) dx.P=L2​∫0L/4​sin2(Lπx​)dx.

Using the trigonometric identity sin⁡2θ=1−cos⁡(2θ)2\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}sin2θ=21−cos(2θ)​:

P=2L∫0L/41−cos⁡(2πxL)2dx.P = \frac{2}{L} \int_0^{L/4} \frac{1 - \cos\left(\frac{2\pi x}{L}\right)}{2} dx.P=L2​∫0L/4​21−cos(L2πx​)​dx.

Simplifying:

P=1L[∫0L/4dx−∫0L/4cos⁡(2πxL)dx].P = \frac{1}{L} \left[\int_0^{L/4} dx - \int_0^{L/4} \cos\left(\frac{2\pi x}{L}\right) dx\right].P=L1​[∫0L/4​dx−∫0L/4​cos(L2πx​)dx].

Evaluating each term:

∫0L/4dx=L4,∫0L/4cos⁡(2πxL)dx=L2πsin⁡(2πxL)∣0L/4.\int_0^{L/4} dx = \frac{L}{4}, \quad \int_0^{L/4} \cos\left(\frac{2\pi x}{L}\right) dx = \frac{L}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \bigg|_0^{L/4}.∫0L/4​dx=4L​,∫0L/4​cos(L2πx​)dx=2πL​sin(L2πx​)​0L/4​.

Substituting:

P=1L[L4−L2πsin⁡(π2)].P = \frac{1}{L} \left[\frac{L}{4} - \frac{L}{2\pi} \sin\left(\frac{\pi}{2}\right)\right].P=L1​[4L​−2πL​sin(2π​)].P=1L[L4−L2π]=14−12π≈0.159.P = \frac{1}{L} \left[\frac{L}{4} - \frac{L}{2\pi}\right] = \frac{1}{4} - \frac{1}{2\pi} \approx 0.159.P=L1​[4L​−2πL​]=41​−2π1​≈0.159.

(b) Ground State EnergyThe energy of the particle in the ground state is given by:

E1=ℏ2π22mL2.E_1 = \frac{\hbar^2 \pi^2}{2mL^2}.E1​=2mL2ℏ2π2​.

Conclusion

These are just two examples of how our expert team tackles complex physics problems with precision and clarity. Whether you're grappling with electromagnetic waves or delving into quantum mechanics, we’ve got you covered. With LiveExamHelper.com, you can rest assured that your exams are in expert hands. Don’t let stress hold you back—hire someone to take my exam today and achieve the grades you’ve always wanted.