Electromagnetic Waves

JEE Main level, single correct, medium

132 of 165

Question 132 of 165

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

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

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

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