Question: 2: The magnetic field of a plane electromagnetic wave is given by \overrightarrow{\mathrm{B}}=3 \times 10^{-8} \cos \left(1.6 \times 10^{3} \mathrm{x}+48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{j}}, then the associated electric field will be :
(1) 3 \times 10^{-8} \cos \left(1.6 \times 10^{3} \mathrm{x}+48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{i}} \mathrm{V} / \mathrm{m}
(2) 3 \times 10^{-8} \sin \left(1.6 \times 10^{3} \mathrm{x}+48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{i}} \mathrm{V} / \mathrm{m}
(3) 9 \sin \left(1.6 \times 10^{3} \mathrm{x}-48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}
(4) 9 \cos \left(1.6 \times 10^{3} \mathrm{x}+48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}
Answer: Option (4)
Explanation:
In a plane electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular.
The magnitude of the electric field is related to the magnetic field by E_0=cB_0, where c=3\times10^8 m/s.
The given magnetic field amplitude is B_0=3\times10^{-8} T.
Therefore, the electric field amplitude is E_0=3\times10^8\times3\times10^{-8}=9 V/m.
The magnetic field is along the \hat{j} direction.
The phase (kx+\omega t) indicates propagation along the negative x-direction.
Using the right-hand rule, the electric field must be along the \hat{k} direction.
The electric and magnetic fields in an electromagnetic wave are in phase, so the electric field also has a cosine dependence.
Hence, the associated electric field is 9 \cos \left(1.6 \times 10^{3} \mathrm{x}+48 \times 10^{10} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}.