Question 31: In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0 \times 10^{10} \mathrm{~Hz} and amplitude 48 \mathrm{Vm}^{-1}. Then the amplitude of oscillating magnetic field is : (Speed of light in free space =3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1} )
(1) 1.6 \times 10^{-7} \mathrm{~T}
(2) 1.6 \times 10^{-6} \mathrm{~T}
(3) 1.6 \times 10^{-9} \mathrm{~T}
(4) 1.6 \times 10^{-8} \mathrm{~T}
Answer: Option (4)
Explanation:
In a plane electromagnetic wave travelling in free space, the amplitudes of electric field and
magnetic field are related by the equation E_{0}=c B_{0}.
Here, the amplitude of electric field is E_{0}=48 \mathrm{~V m^{-1}}
and the speed of light is c=3 \times 10^{8} \mathrm{~m s^{-1}}.
The amplitude of the magnetic field is given by B_{0}=\frac{E_{0}}{c}.
Substituting the given values, B_{0}=\frac{48}{3 \times 10^{8}}.
On simplification, B_{0}=16 \times 10^{-8} \mathrm{~T}.
This can be written as
1.6 \times 10^{-7} \times 10^{-1} \mathrm{~T}=1.6 \times 10^{-8} \mathrm{~T}.
Hence, the correct answer is Option (4).