Electromagnetic WavesNEET Physics · Class 12 · NCERT Chapter 8

Solution

Order: gamma < X < UV < visible < IR < microwave < radio. Radio has the longest λ.

Solution

$E_0=c{B}_0$, so $E_0/B_0=c$.

Solution

E and B are mutually perpendicular and oscillate in phase, both perpendicular to direction of propagation.

Solution

$I_D={\epsilon }_{0}d{\Phi }_E/dt$. Needs a CHANGING electric flux.

Solution

$c=1/\sqrt{{\epsilon }_0{\mu }_0}$.

Solution

Gamma rays > X-rays > UV > visible.

Solution

$B_0=E_0/c=60/(3\times 10^8)=2\times 10^{-7}\text{T}$.

Solution

UV kills bacteria; used in OT lamps and water sterilisation.

Solution

Gamma > X-ray. Higher photon energy means deeper penetration.

Solution

$u_{\text{avg}}=\frac{1}{2}{\epsilon }_0{E}_0^2$ (using $E_{\text{rms}}^2=E_0^2/2$, then $u={\epsilon }_0{E}_{\text{rms}}^2$).

Solution

Absorber: $P=I/c$. Reflector: $P=2I/c$.

Solution

Reflector: momentum reverses, so $P=2I/c$.

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Frequency stays, but $v=c/n$ and $\lambda ={\lambda }_0/n$.

Solution

No conduction current between plates. Only displacement current is present, and it produces the B-field.

Solution

2.45 GHz is the resonance frequency for water rotation.

Solution

$u_E=u_B$. The two carry equal energy density.

Solution

$E_0=c{B}_0=3\times 10^8\times 1.5\times 10^{-7}=45\text{V/m}$.

Solution

Thermopiles and bolometers detect IR via heating.

Solution

Time-varying E gives displacement current; time-varying B gives induced EMF (Faraday).

Solution

$n=\sqrt{{\epsilon }_r}=2$ for ${\mu }_r=1$. $v=c/n=c/2$.

Solution

$I=uc$.

Solution

Only ACCELERATING charges radiate EM waves. A constant current (steady drift) radiates none.

Solution

Radio waves reflect off the ionosphere → long-distance comms.

Solution

$\lambda =c/f=3\times 10^8/(7.5\times 10^{14})=4\times 10^{-7}\text{m}=400\text{nm}$ (violet).

Solution

$P=I/c=1361/(3\times 10^8)\approx 4.5\times 10^{-6}\text{Pa}$.

Solution

TV remotes use IR LEDs at around 940 nm.

Solution

$I_D=CdV/dt=4\times 10^{-6}\times 5\times 10^4=0.2\text{A}$.

Solution

$u=\frac{1}{2}{\epsilon }_0{E}_0^2=0.5\times 8.85\times 10^{-12}\times 9\times 10^4\approx 3.98\times 10^{-7}$ J/m³.

Solution

$10^{-10}$ m = 0.1 nm. X-rays are 0.01 to 10 nm.

Solution

Sun's visible light heats the Earth, which radiates IR. Greenhouse gases absorb the outgoing IR.