JEE Main2012PhysicsElectromagnetic WavesActual
A radio transmitter transmits at 830 kHz . At a certain distance from the transmitter magnetic field has amplitude 4.82 10⁻¹¹ ~T . The electric field and the wavelength are respectively
Options
- A0.014 ~N / C , 36 ~m
- B0.14 ~N / C , 36 ~m
- C0.14 ~N / C , 360 ~m
- D0.014 ~N / C , 360 ~m
Correct answer
D. 0.014 ~N / C , 360 ~m
Step-by-step solution
Frequency of EM wave v=830 KHz =830 10^3 ~Hz . Magnetic field, B=4.82 10⁻¹¹ ~T As we know, frequency, v= c aligned or & = c v = 3 10^8 830 10^3 & 360 ~m And, E & =B C=4.82 10⁻¹¹ 3 10^8 & =0.014 ~N / C aligned