Q.46. An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, mho

Q: What is the correct answer for JEE Main Physics Q46?

The correct answer is 1800. Frequency $f = 100$ MHz $= 100 \times 10^6$ Hz $= 10^8$ Hz. Angular frequency $\omega = 2\pi f = 2\pi \times 10^8$ rad/s. Conductivity $\sigma = 10$ mho/m. The ratio of maximum conducting current density ($J_c$) to maximum displacement current density ($J_d$) is given by $\frac{J_c}{J_d} = \frac{\sigma E_0}{\omega \epsilon E_0} = \frac{\sigma}{\omega \epsilon}$. Here, $\epsilon = \epsilon_0$ for a

Q: Q.46. An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, mho/m. The ratio of maximum conducting current density to maximum displacement current density is…

The correct answer is 1800. Frequency $f = 100$ MHz $= 100 \times 10^6$ Hz $= 10^8$ Hz. Angular frequency $\omega = 2\pi f = 2\pi \times 10^8$ rad/s. Conductivity $\sigma = 10$ mho/m. The ratio of maximum conducting current density ($J_c$) to maximum displacement current density ($J_d$) is given by $\frac{J_c}{J_d} = \frac{\sigma E_0}{\omega \epsilon E_0} = \frac{\sigma}{\omega \epsilon}$. Here, $\epsilon = \epsilon_0$ for a

Question

Q.46. An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity,  mho/m. The ratio of maximum conducting current density to maximum displacement current density is ________. [Take  Nm/C]

FirstInTest · Accepted Answer

Frequency $f = 100$ MHz $= 100 	imes 10^6$ Hz $= 10^8$ Hz. Angular frequency $\omega = 2\pi f = 2\pi 	imes 10^8$ rad/s. Conductivity $\sigma = 10$ mho/m. The ratio of maximum conducting current density ($J_c$) to maximum displacement current density ($J_d$) is given by $\frac{J_c}{J_d} = \frac{\sigma E_0}{\omega \epsilon E_0} = \frac{\sigma}{\omega \epsilon}$. Here, $\epsilon = \epsilon_0$ for a non-magnetic medium. $\epsilon_0 = \frac{1}{4\pi k}$ where $k = 9 	imes 10^9$ Nm$^2$/C$^2$. So, $\epsilon_0 = \frac{1}{4\pi 	imes 9 	imes 10^9}$ F/m. $\frac{J_c}{J_d} = \frac{\sigma}{\omega \epsilon_0} = \frac{10}{(2\pi 	imes 10^8) 	imes \frac{1}{4\pi 	imes 9 	imes 10^9}} = \frac{10 	imes 4\pi 	imes 9 	imes 10^9}{2\pi 	imes 10^8} = \frac{10 	imes 2 	imes 9 	imes 10^9}{10^8} = 10 	imes 18 	imes 10 = 1800$.

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