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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2012 · 26 May · Online · Physics · Electromagnetic Waves

A radio transmitter transmits at 830 kHz . At a certain distance...

A radio transmitter transmits at 830kHz830 \mathrm{kHz}. At a certain distance from the transmitter magnetic field has amplitude 4.82×10−11 T4.82 \times 10^{-11} \mathrm{~T}. The electric field and the wavelength are respectively

Options

  1. A0.014 N/C,36 m0.014 \mathrm{~N} / \mathrm{C}, 36 \mathrm{~m}
  2. B0.14 N/C,36 m0.14 \mathrm{~N} / \mathrm{C}, 36 \mathrm{~m}
  3. C0.14 N/C,360 m0.14 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}
  4. D0.014 N/C,360 m0.014 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}

Correct answer

D. 0.014 N/C,360 m0.014 \mathrm{~N} / \mathrm{C}, 360 \mathrm{~m}

Step-by-step solution

Frequency of EM wave v=830KHz=830×103Hzv=830 \mathrm{KHz}=830 \times 10^3 \mathrm{ Hz}. Magnetic field, B=4.82×10−11TB=4.82 \times 10^{-11} \mathrm{ T} As we know, frequency, v=cλv=\frac{c}{\lambda} or λ=cv=3×108830×103λ≃360m And, E=BC=4.82×10−11×3×108=0.014N/C\begin{aligned}\text{ or } \lambda& =\frac{c}{v}=\frac{3 \times 10^8}{830 \times 10^3}\\ \lambda & \simeq 360 \mathrm{ m}\\ \text{ And, } E & =B C=4.82 \times10^{-11} \times3 \times10^8 \\ & =0.014 \mathrm{ N}/ \mathrm{C} \end{aligned}$$
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