Electromagnetic Waves Questions and Answers

Answer: B

The crystal is Galena.

Galena, natural lead sulphide (PbS), was the crystal in the early 'cat's whisker' detectors of radio receivers. A fine wire touching the crystal formed a point-contact rectifier, one of the first semiconductor devices, which demodulated the radio signal.

Silicon and germanium are semiconductors used in later diodes, but they are refined materials, not naturally occurring crystals used in that role. Ruby is the crystal of the first laser, and zinc selenide is an infrared optical material.

Answer: C

The time period is \(5 \times 10^{-7}s\).

The wave oscillates at the same frequency as the accelerated charge that produces it, so

\[ \begin{aligned} T &= \frac{1}{f} = \frac{1}{2 \times 10^{6}} \\ &= 5 \times 10^{-7}\ \text{s} \end{aligned} \]

\(2 \times 10^{-7}\) s confuses the period with the frequency's digits, and \(2.5 \times 10^{-7}\) s halves the correct value.

Answer: A

The correct statement is that they can travel through vacuum.

Electromagnetic waves consist of oscillating electric and magnetic fields that sustain each other, so they need no material medium; light reaches us from the sun through the vacuum of space.

They contain both fields together, perpendicular to each other and to the direction of travel, so 'only electric' and 'only magnetic' are wrong, and so is 'require a medium'. Their speed in vacuum is \(3 \times 10^{8}\) m s^-1; the last option gives the unit as cm s^-1, which is a hundred times too small.

Answer: A

A is a microwave and B is visible light.

Microwaves have wavelengths from about 1 mm to about 30 cm, so 2 cm is a microwave (radar and microwave ovens work in this range). Visible light runs from about 400 nm to 700 nm, so 580 nm is visible, yellow light.

Infrared lies between about 700 nm and 1 mm, too short for 2 cm. Radio waves are longer still, from tens of centimetres upward. Ultraviolet lies below 400 nm, too short for 580 nm.

Answer: C

The wavelength band is 40m-25m.

\(\lambda = \frac{c}{f}\) with \(c = 3 \times 10^{8}\) m/s:

\[ \begin{aligned} \lambda_{1} &= \frac{3 \times 10^{8}}{7.5 \times 10^{6}} = 40\ \text{m} \\ \lambda_{2} &= \frac{3 \times 10^{8}}{12 \times 10^{6}} = 25\ \text{m} \end{aligned} \]

Higher frequency means shorter wavelength, so the band runs from 40 m down to 25 m.

Answer: A

The symmetrical counterpart is the Ampere-Maxwell law.

Faraday's law says a changing magnetic flux produces an electric field (an induced emf). Maxwell's addition to Ampere's law, the displacement current, says a changing electric flux produces a magnetic field:

\[ \begin{aligned} &\oint \vec{B}\cdot d\vec{l} \\ &= \mu_0\left(I + \varepsilon_0\frac{d\Phi_E}{dt}\right) \end{aligned} \]

The two laws mirror each other, which is what allows electromagnetic waves. Ampere's circuital law without Maxwell's term has no changing-field part; Gauss's law and Coulomb's law concern static charges, and Lenz's law only gives the direction of the induced current.

Answer: C

The speed is \(\frac{c}{2}\).

The speed of an electromagnetic wave in a medium is \(v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_r\varepsilon_r}}\).

\[ v = \frac{c}{\sqrt{1 \times 4}} = \frac{c}{2} \]

\(\frac{c}{4}\) forgets the square root, and 2c would mean the wave travels faster than in a vacuum, which a material medium does not allow. The refractive index of this medium is 2.

Answer: E

The key gives \(1.2 \times 10^{-8}\) Nm^-2.

For a completely absorbing surface, the momentum delivered per second is the energy delivered per second divided by \(c\). Reading the 3.6 J per unit time as falling on unit area (an intensity of 3.6 W m^-2), the pressure is

\[ \begin{aligned} P &= \frac{I}{c} \\ &= \frac{3.6}{3 \times 10^{8}} \\ &= 1.2 \times 10^{-8}\ \text{Nm}^{-2} \end{aligned} \]

The other options multiply by \(c\) instead of dividing, which gives enormous pressures. A perfectly reflecting surface would receive twice this pressure.

Answer: C

Microwaves are short wavelength radio waves.

In the electromagnetic spectrum, microwaves (wavelengths of about 1 mm to 0.1 m, frequencies of roughly 1 to 300 GHz) lie at the short-wavelength end of the radio region. They are used in radar, microwave ovens and satellite links.

The 54 to 890 MHz band and ordinary radio and television broadcasting belong to longer radio waves. Hot bodies and molecular vibrations produce infrared radiation, and ordinary glass absorbs ultraviolet, not microwaves.

Answer: B

The magnetic field amplitude is \(1.5 \times 10^{-7}\) T.

In an electromagnetic wave the field amplitudes are related by \(E_0 = cB_0\).

\[ \begin{aligned} B_0 &= \frac{45}{3 \times 10^{8}} \\ &= 1.5 \times 10^{-7}\ \text{T} \end{aligned} \]

\(1.5 \times 10^{-8}\) T is the same digits with the power of ten slipped by one. The ratio \(\frac{E_0}{B_0} = c\) holds for any electromagnetic wave in vacuum, whatever its frequency.