Question

The magnetic field of a plane electromagnetic wave is given by $\vec{B}=3\times 10^{-8}\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)\hat{j}$, then the associated electric field will be :

• A. $3\times 10^{-8}\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)\hat{i}V/m$
• B. $3\times 10^{-8}\sin \left(1.6\times 10^{3}x+48\times 10^{10}t\right)\hat{i}V/m$
• C. $9\sin \left(1.6\times 10^{3}x-48\times 10^{10}t\right)\hat{k}V/m$
• D. $9\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)\hat{k}V/m$

Answer 1

Answer: (D)

$B=3\times 10^{-8}\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)$
$C=\frac{\omega }{k}=\frac{48\times 10^{10}}{1.6\times 10^3}=3\times 10^{8}m/s$
$C=E_0/B_0$
$E=3\times 10^{-8}\times 3\times 10^8=9N/C$
$\therefore E=9\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)$