JEE Main & Advanced Physics EM Waves JEE PYQ-Electro Magnetic Waves

A radio transmitter transmits at 830 kHz. At a certain distance from the transmitter magnetic field has amplitude \[4.82\times {{10}^{-11}}T.\]The electric field and the wavelength are respectively[JEE ONLINE 26-05-2012]

A) 0.014N/C, 36m

B) 0.14N/C, 36m

C) 0.14N/C, 360m

D) 0.014N/C, 360m

Correct Answer: D

Solution :

[d] Frequency of EM wave \[\upsilon =830\,KHz\] \[=830\times {{10}^{3}}Hz.\]

Magnetic field, \[B=4.82\times {{10}^{-11}}T\]

As we know, frequency, \[\upsilon =\frac{c}{\lambda }\]

or\[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{830\times {{10}^{3}}}\]\[\lambda \simeq 360m\]

And, \[E=BC=4.82\times {{10}^{-11}}\times 3\times {{10}^{8}}\]

= 0.014 N/C