A radio transmitter transmits at 830 , kHz. A | vedclass

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(D) Given frequency $v = 830 \, kHz = 830 	imes 10^3 \, Hz$.Amplitude of magnetic field $B_0 = 4.82 	imes 10^{-11} \, T$.Speed of light $c = 3 	ime

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$0.014 \, N/C, 360 \, m$

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(D) Given frequency $v = 830 \, kHz = 830 	imes 10^3 \, Hz$.Amplitude of magnetic field $B_0 = 4.82 	imes 10^{-11} \, T$.Speed of light $c = 3 	imes 10^8 \, m/s$.$1$. Calculation of wavelength $(\lambda)$:$\lambda = \frac{c}{v} = \frac{3 	imes 10^8}{830 	imes 10^3} = \frac{3000}{83} \approx 361.4 \, m \approx 360 \, m$.$2$. Calculation of electric field amplitude $(E_0)$:Using the relation $E_0 = B_0 c$,$E_0 = (4.82 	imes 10^{-11} \, T) 	imes (3 	imes 10^8 \, m/s)$,$E_0 = 14.46 	imes 10^{-3} \, N/C \approx 0.014 \, N/C$.Thus,the electric field is $0.014 \, N/C$ and the wavelength is $360 \, m$.

$A$ radio transmitter transmits at $830 \, kHz$. At a certain distance from the transmitter,the magnetic field has an amplitude of $4.82 \times 10^{-11} \, T$. The electric field and the wavelength are respectively:

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