A laser beam has an intensity of 4.0 times 10 | vedclass

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(D) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 E_0^2 c$. From this,the amplitude of the electric field $E_0$

Vedclass · Accepted Answer

$1.83$

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(D) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 E_0^2 c$. From this,the amplitude of the electric field $E_0$ is $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$. We know that the relation between the amplitudes of the electric and magnetic fields is $E_0 = B_0 c$,which implies $B_0 = \frac{E_0}{c}$. Substituting $E_0$,we get $B_0 = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2I}{\epsilon_0 c^3}}$. Given $I = 4.0 	imes 10^{14} \ W/m^2$,$\epsilon_0 = 8.85 	imes 10^{-12} \ C^2/Nm^2$,and $c = 3 	imes 10^8 \ m/s$: $B_0 = \sqrt{\frac{2 	imes 4.0 	imes 10^{14}}{8.85 	imes 10^{-12} 	imes (3 	imes 10^8)^3}} = \sqrt{\frac{8.0 	imes 10^{14}}{8.85 	imes 10^{-12} 	imes 27 	imes 10^{24}}} = \sqrt{\frac{8.0 	imes 10^{14}}{238.95 	imes 10^{12}}} = \sqrt{\frac{800}{238.95}} \approx \sqrt{3.348} \approx 1.83 \ T$.

$A$ laser beam has an intensity of $4.0 \times 10^{14} \ W/m^{2}$. The amplitude of the magnetic field associated with the beam is . . . . . . $T$. (Take $\epsilon_{0} = 8.85 \times 10^{-12} \ C^{2}/Nm^{2}$ and $c = 3 \times 10^{8} \ m/s$)

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