Electromagnetic Waves | Physics JEE Main questions with Solutions

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Topic Question Set

Q 11 :
A parallel plate capacitor has a capacitance $C = 200$ $p F$ . It is connected to 230 V AC supply with an angular frequency 300 rad/s. The RMS value of conduction current in the circuit and displacement current in the capacitor respectively are [2024]

$1.38 μ A$ and $1.38 μ A$
$14.3 μ A$ and $143 μ A$
$13.8 μ A$ and $138 μ A$
$13.8 μ A$ and $13.8 μ A$

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(D)

Q 12 :
The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) {NC}^{- 1}$ . The magnetic field induction of this wave is (in SI unit): [2024]

$\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})$
$\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})$
$\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})$
$\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})$

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(4)

$\vec{E} = 40 \cos (ω t - \frac{ω}{c} z) {NC}^{- 1} \hat{i}$

$\vec{E}$ is along $+ x$ direction

$\vec{v}$ is along $+ z$ direction

and Direction of $\hat{c}$ must be along $\vec{E} \times \vec{B}$ .

So direction of $\vec{B}$ will be along $+ y$ and magnitude of $\vec{B}$ will be $\frac{E_{0}}{c} .$

$B_{0} = \frac{E_{0}}{c} = \frac{40}{c}$

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