Frequency of EM wave v=830KHz =830×103 Hz. Magnetic field, B=4.82×10−11 T As we know, frequency, v=λc or λλ And, E=vc=830×1033×108≃360 m=BC=4.82×10−11×3×108=0.014 N/C
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…
JEE Main 2012 — Physics Electromagnetism
A radio transmitter transmits at 830kHz. At a certain distance from the transmitter magnetic field has amplitude 4.82×10−11 T. The electric field and the wavelength are respectively
Held on 26 May 2012 · Verified 6 Jul 2026.
Options
- A
0.014 N/C,36 m
- B
0.14 N/C,36 m
- C
0.14 N/C,360 m
- D
0.014 N/C,360 m
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Electromagnetism in other years
More JEE Main Electromagnetism questions
A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{j}$. Its associated electric field vector is ______.
A plane electromagnetic wave is moving in free space with velocity $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\hat{j}$ is the unit vector along $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:
A charged particle moves in a uniform magnetic field. If the velocity is perpendicular to the field, the path of the particle is:
A monochromatic source of light operating at $15$ kW emits $2.5\times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h=6.6\times 10^{-34}$ J·s and $c=3\times 10^8$ m/s).
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10 \mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$. $\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]$
Other Physics topics for JEE Main
Keep practicing Electromagnetism
Work through every JEE Main Electromagnetism PYQ, year by year.