Question: 9. The electric field in a plane electromagnetic wave is given by E_{z}=60 \cos \left(5 x+1.5 \times 10^{9} t\right)\,V / m. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field)
(1) B_{y}=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^{9} t\right)\,T
(2) B_{x}=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^{9} t\right)\,T
(3) B_{z}=60 \cos \left(5 x+1.5 \times 10^{9} t\right)\,T
(4) B_{y}=60 \sin \left(5 x+1.5 \times 10^{9} t\right)\,T
Answer: Option (1)
Explanation:
In an electromagnetic wave, electric field \vec{E}, magnetic field \vec{B} and direction of propagation are mutually perpendicular.
The given electric field is
\vec{E}=E_z \hat{z}=60\cos(5x+1.5\times 10^9 t)\,\hat{z}\;V/m.The argument (5x+1.5\times 10^9 t) indicates the wave is travelling along the negative x-direction or simply depends on x. Thus, propagation is along the x-axis.
Since \vec{E} \perp \vec{B} \perp \text{direction of propagation}, and direction of propagation = x-axis, electric field = z-axis, therefore magnetic field must be along the y-axis.
The magnitudes satisfy the relation
\dfrac{E_0}{B_0}=cwhere c=3\times 10^8\,m/s.
So,
B_0=\dfrac{E_0}{c}=\dfrac{60}{3\times 10^8}=2\times 10^{-7}\,T.Thus, the magnetic field is
B_y = 2\times 10^{-7}\cos(5x+1.5\times 10^9 t)\,T.This matches Option (1).