IMO Practice Test: Electromagnetic Waves | Class 12 Physics

Question 1 of 12

A parallel-plate capacitor has plate area $A$ and the field between the plates grows at $\frac{dE}{dt}$. The displacement current equals:

$\epsilon_0\frac{dE}{dt}$

$\epsilon_0 A\frac{dE}{dt}$

$A\frac{dE}{dt}$

$\mu_0 A\frac{dE}{dt}$

Explanation: $I_d=\epsilon_0\frac{d\Phi_E}{dt}=\epsilon_0 A\frac{dE}{dt}$, since $\Phi_E=EA$.

Question 2 of 12

If the permittivity of a medium is $4\epsilon_0$ and permeability $\mu_0$, the wave speed in it is:

$2c$

$\frac{c}{2}$

$c$

$4c$

Explanation: $v=\frac{1}{\sqrt{\mu_0(4\epsilon_0)}}=\frac{1}{2}\cdot\frac{1}{\sqrt{\mu_0\epsilon_0}}=\frac{c}{2}$.

Question 3 of 12

An EM wave has intensity $I$. If its electric field amplitude is doubled, the intensity becomes:

$2I$

$4I$

$I$

$8I$

Explanation: $I=\frac{1}{2}\epsilon_0 E_0^2 c\propto E_0^2$, so doubling $E_0$ quadruples the intensity.

Question 4 of 12

A 60 W source radiates uniformly. Treating it as a point source, the intensity at 2 m (assuming all power is EM radiation) is about:

$1.2\ \text{W/m}^2$

$15\ \text{W/m}^2$

$4.8\ \text{W/m}^2$

$0.6\ \text{W/m}^2$

Explanation: $I=\frac{P}{4\pi r^2}=\frac{60}{4\pi(2)^2}=\frac{60}{50.3}\approx1.2\ \text{W/m}^2$.

Question 5 of 12

For an EM wave, the average total energy density in terms of $E_0$ is:

$\frac{1}{2}\epsilon_0 E_0^2$

$\frac{1}{4}\epsilon_0 E_0^2$

$\epsilon_0 E_0^2$

$2\epsilon_0 E_0^2$

Explanation: Total $u=\epsilon_0 E^2$; averaging $E^2$ over a cycle gives $\frac{E_0^2}{2}$, so $\langle u\rangle=\frac{1}{2}\epsilon_0 E_0^2$.

Question 6 of 12

Light of intensity $I$ falls on a perfectly reflecting mirror. The radiation pressure is:

$\frac{I}{c}$

$\frac{2I}{c}$

$\frac{I}{2c}$

$Ic$

Explanation: A reflector reverses the photon momentum, so it receives twice as much: $P=\frac{2I}{c}$.

Question 7 of 12

The frequency of a $0.3\ \text{nm}$ X-ray is ($c=3\times10^{8}$):

$1\times10^{18}\ \text{Hz}$

$1\times10^{15}\ \text{Hz}$

$1\times10^{21}\ \text{Hz}$

$3\times10^{17}\ \text{Hz}$

Explanation: $f=\frac{c}{\lambda}=\frac{3\times10^{8}}{0.3\times10^{-9}}=1\times10^{18}\ \text{Hz}$.

Question 8 of 12

Two EM waves have wavelengths in the ratio 1:3. The ratio of their photon energies is:

1:3

3:1

1:9

9:1

Explanation: $E=hf=\frac{hc}{\lambda}\propto\frac{1}{\lambda}$, so the energy ratio is the inverse, 3:1.

Question 9 of 12

In a charging capacitor with $I=3\ \text{A}$, the magnetic field between the plates is produced by:

zero current

the conduction current only

the displacement current $I_d=3\ \text{A}$

twice the conduction current

Explanation: Between the plates the magnetic field is produced by the displacement current, equal to the 3 A conduction current.

Question 10 of 12

An EM wave travels along $+z$ with $\vec{E}$ along $+x$. The magnetic field $\vec{B}$ points along:

$+x$

$+z$

$+y$

$-z$

Explanation: Propagation is along $\vec{E}\times\vec{B}$; with $\vec{E}$ along $+x$ and travel along $+z$, $\vec{B}$ must be along $+y$ ($\hat{x}\times\hat{y}=\hat{z}$).

Question 11 of 12

The Sun's radiation pressure on a perfectly absorbing panel where $I=1360\ \text{W/m}^2$ is about:

$4.5\times10^{-6}\ \text{Pa}$

$4.5\times10^{-3}\ \text{Pa}$

$1360\ \text{Pa}$

$9\times10^{-6}\ \text{Pa}$

Explanation: $P=\frac{I}{c}=\frac{1360}{3\times10^{8}}\approx4.5\times10^{-6}\ \text{Pa}$.

Question 12 of 12

A wave has $\lambda=10\ \text{m}$ in vacuum. Identify the band and its frequency:

radio, $3\times10^{7}\ \text{Hz}$

microwave, $3\times10^{10}\ \text{Hz}$

infrared, $3\times10^{13}\ \text{Hz}$

X-ray, $3\times10^{16}\ \text{Hz}$

Explanation: $f=\frac{c}{\lambda}=\frac{3\times10^{8}}{10}=3\times10^{7}\ \text{Hz}$, which is a radio wave.

IMO Practice Test — Electromagnetic Waves