An electromagnetic wave travelling in the $x-$ direction has frequency of $2 \times 10^{14}\,Hz$ and electric field amplitude of $27\,Vm^{-1}$ . From the options given below, which one describes the magnetic field for this wave ?
$\vec B\,(x\,,\,t) = (3 \times {10^{ - 8}}\,T)\,\hat j \;\sin \,[2\pi \,(1.5 \times {10^{ - 8}}\,x\, - \,2 \times {10^{14}}\,t)]$
$\vec B\,(x\,,\,t) = (9 \times {10^{ - 8}}\,T)\,\hat i\; \sin \,[2\pi \,(1.5 \times {10^{ - 8}}\,x\, - \,2 \times {10^{14}}\,t)]$
$\vec B\,(x\,,\,t) = (9 \times {10^{ - 8}}\,T)\,\hat j\;\sin \,[(1.5 \times {10^{ - 6}}\,x\, - \,2 \times {10^{14}}\,t)]$
$\vec B\,(x\,,\,t) = (9 \times {10^{ - 8}}\,T)\,\hat k \;\sin \,[2\pi \,(1.5 \times {10^{ - 6}}\,x\, - \,2 \times {10^{14}}\,t)]$
Light wave traveling in air along $x$-direction is given by $E _{ y }=540 \sin \pi \times 10^{4}( x - ct ) Vm ^{-1}$. Then, the peak value of magnetic field of wave will be $\dots \times 10^{-7}\,T$ (Given $c =3 \times 10^{8}\,ms ^{-1}$ )
About $5 \%$ of the power of a $100\; W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation
$(a)$ at a distance of $1 \;m$ from the bulb?
$(b)$ at a distance of $10\; m ?$ Assume that the radiation is emitted isotropically and neglect reflection.
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric $\left( U _{ e }\right)$ and magnetic $\left( U _{ m }\right)$ fields is
The magnetic field of an electromagnetic wave is given by
$\vec B = 1.6 \times {10^{ - 6}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\hat i + \hat j} \right)\frac{{Wb}}{{{m^2}}}$ The associated electric field will be
A plane electromagnetic wave is travelling in the positive $X-$axis. At the instant shown electric field at the extremely narrow dashed rectangle is in the $-ve$ $z$ direction and its magnitude is increasing. Which diagram correctly shows the direction and relative magnitudes of magnetic field at the edges of rectangle :-