"Electromagnetic Waves" — Practice

Question 1

A parallel-plate capacitor with circular plates of radius $5$ cm is being charged so that the electric field between the plates increases at $5	imes10^{11}$ V m$^{-1}$ s$^{-1}$. Find the displacement current and the magnetic field at a radius of $2$ cm from the axis.

Accepted Answer

$A=\pi\left(0.05ight)^{2}=7.85	imes10^{-3}$ m$^{2}$

$i_d=\varepsilon_0A\dfrac{dE}{dt}=\left(8.85	imes10^{-12}ight)\left(7.85	imes10^{-3}ight)\left(5	imes10^{11}ight)=0.0347$ A

$B=\dfrac{\mu_0i_dr}{2\pi R^{2}}=\dfrac{\left(2	imes10^{-7}ight)\left(0.0347ight)\left(0.02ight)}{\left(0.05ight)^{2}}$

$B=5.6	imes10^{-8}$ T

Question 2

Explain why the magnetic field just outside the edge of a charging capacitor's plates equals that just outside the connecting wire at the same radius.

Accepted Answer

For a loop of radius $r>R$ around the wire, the enclosed conduction current is $i_c$.

For the same loop placed at the capacitor, the enclosed conduction current is zero but the enclosed displacement current is $i_d$.

Since $i_d=i_c$ exactly, the right-hand side of the Ampère-Maxwell law is identical in both cases, and so is $B=\dfrac{\mu_0i}{2\pi r}$.

The current is continuous through the gap; only its form changes.

Question 3

A non-magnetic material has relative permittivity $4$. Find its refractive index, the speed of light in it, and the wavelength there of $500$ nm vacuum light.

Accepted Answer

$n=\sqrt{\varepsilon_r}=\sqrt4=2$

$v=\dfrac{c}{n}=1.5	imes10^{8}$ m s$^{-1}$

$\lambda'=\dfrac{500}{2}=250$ nm

The frequency is unchanged at $6	imes10^{14}$ Hz, since the fields on the two sides of the boundary must stay in step.

Question 4

An electromagnetic wave travels along $-\hat k$ with its electric field along $+\hat i$ and peak value $60$ V m$^{-1}$. Find the direction and peak value of the magnetic field.

Accepted Answer

We need $\hat i	imes\hat B=-\hat k$.

Since $\hat i	imes\hat j=\hat k$ and $\hat i	imes\left(-\hat jight)=-\hat k$, the magnetic field is along $-\hat j$.

$B_0=\dfrac{E_0}{c}=\dfrac{60}{3	imes10^{8}}=2	imes10^{-7}$ T

Both fields peak at the same instant, since they are in phase.

Question 5

A wave has peak electric field $30$ V m$^{-1}$. Find the average energy density, the intensity, and verify that the electric and magnetic contributions are equal.

Accepted Answer

$\bar u=	frac12\varepsilon_0E_0^{2}=	frac12\left(8.85	imes10^{-12}ight)(900)=3.98	imes10^{-9}$ J m$^{-3}$

$I=\bar u\,c=3.98	imes10^{-9}	imes3	imes10^{8}=1.19$ W m$^{-2}$

Check: $B_0=10^{-7}$ T, so $\dfrac{B_0^{2}}{4\mu_0}=\dfrac{10^{-14}}{4	imes4\pi	imes10^{-7}}=1.99	imes10^{-9}$ J m$^{-3}$, exactly half the total — equal to the electric share.

Question 6

A $60$ W bulb radiates uniformly, converting $5\%$ of its input into visible light. Find the intensity and peak electric field at $2$ m.

Accepted Answer

Radiated power $=0.05	imes60=3$ W

$I=\dfrac{3}{4\pi\left(4ight)}=0.0597$ W m$^{-2}$

$E_0=\sqrt{\dfrac{2I}{\varepsilon_0c}}=\sqrt{\dfrac{0.1194}{2.655	imes10^{-3}}}=\sqrt{45.0}=6.7$ V m$^{-1}$

At $4$ m the intensity would be a quarter of this but the field only half, since $E_0\propto1/r$.

Question 7

A laser beam of power $2$ W and cross-section $1$ mm$^{2}$ falls normally on a perfectly reflecting mirror. Find the radiation pressure and the force, and state how each changes if the beam is focused to $0.1$ mm$^{2}$.

Accepted Answer

$I=\dfrac{2}{10^{-6}}=2	imes10^{6}$ W m$^{-2}$

$P=\dfrac{2I}{c}=\dfrac{4	imes10^{6}}{3	imes10^{8}}=1.3	imes10^{-2}$ Pa

$F=\dfrac{2	imes	ext{power}}{c}=\dfrac{4}{3	imes10^{8}}=1.3	imes10^{-8}$ N

Focusing to a tenth of the area **multiplies the pressure by ten** but leaves the force completely unchanged.

Question 8

A solar panel of area $2$ m$^{2}$ absorbs sunlight of intensity $1000$ W m$^{-2}$. Find the force on it, and compare with the force if it were a perfect mirror.

Accepted Answer

$F_{absorbed}=\dfrac{IA}{c}=\dfrac{1000	imes2}{3	imes10^{8}}=6.7	imes10^{-6}$ N

$F_{reflected}=2	imes6.7	imes10^{-6}=1.3	imes10^{-5}$ N

Both are minute, but a solar sail of ten thousand square metres would feel about a tenth of a newton continuously and without any fuel.

Question 9

Unpolarised light of intensity $I_0$ falls on two polarisers whose axes are at $60^{\circ}$. Find the emergent intensity. Then find the angle of a third polariser, placed between them, that maximises the final intensity.

Accepted Answer

Two polarisers: $I=\dfrac{I_0}{2}\cos^{2}60^{\circ}=\dfrac{I_0}{8}$

With a third at angle $	heta$ to the first: $I=\dfrac{I_0}{2}\cos^{2}	heta\cos^{2}\left(60^{\circ}-	hetaight)$

This is maximised when the angle is split equally, $	heta=30^{\circ}$:

$I=\dfrac{I_0}{2}\left(0.75ight)\left(0.75ight)=0.281I_0$

more than twice the two-polariser value.

Question 10

Find Brewster's angle for glass of refractive index $1.5$, and the angle of refraction at that incidence.

Accepted Answer

$	an	heta_B=1.5\qquad\Rightarrow\qquad 	heta_B=56.3^{\circ}$

Since the reflected and refracted rays are perpendicular,

$	heta_r=90^{\circ}-56.3^{\circ}=33.7^{\circ}$

Check with Snell's law: $\sin56.3^{\circ}=0.832$ and $1.5\sin33.7^{\circ}=1.5	imes0.555=0.832$, which agrees.

Question 11

Estimate how much more strongly light of wavelength $450$ nm is scattered by air molecules than light of $650$ nm, and explain why sunsets are red.

Accepted Answer

$\dfrac{I_{450}}{I_{650}}=\left(\dfrac{650}{450}ight)^{4}=\left(1.444ight)^{4}=4.35$

Blue is scattered about **four times** as strongly.

Overhead, what reaches the eye is scattered light, which is therefore blue. At sunset the direct beam travels a long slanting path through the atmosphere, losing most of its blue to scattering along the way, so what survives to reach the eye is red.

Question 12

A radar operates at $10$ GHz. Find the wavelength, and state why this band rather than visible light is used for detecting aircraft in cloud.

Accepted Answer

$\lambda=\dfrac{3	imes10^{8}}{10^{10}}=3	imes10^{-2}$ m, that is $3$ cm.

Cloud droplets are a few micrometres across, thousands of times smaller than this wavelength, so they scatter the radar beam negligibly and it passes straight through.

Visible light has a wavelength comparable to the droplet size, so it is scattered strongly — which is exactly why cloud is opaque to the eye and transparent to radar.

Practice — "Electromagnetic Waves"

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