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Question

The electric field of an electromagnetic wave travelling through a medium is given by $\vec{E}(x,t) = 25 \sin(2.0 \times 10^{15} t - 10^7 x) \hat{n}$
then the refractive index of the medium is ________.
(All given measurement are in SI units)

The correct answer is
$1.7$

Key Concept: Refractive Index Calculation

The refractive index ($n$) of a medium is defined as the ratio of the speed of light in vacuum ($c$) to the speed of light in the medium ($v$), i.e., $n = c/v$. The speed of light ($v$) in a medium through which an electromagnetic wave propagates is related to the wave's angular frequency ($\omega$) and wave number ($k$) by the formula $v = \omega/k$.

Extracting Wave Parameters

The electric field of the electromagnetic wave is given by $\vec{E}(x,t) = 25 \sin(2.0 \times 10^{15} t - 10^7 x) \hat{n}$. By comparing this to the general form $\vec{E}(x,t) = E_0 \sin(\omega t - kx)$, we can identify the parameters:

  • Angular frequency, $\omega = 2.0 \times 10^{15}$ rad/s.
  • Wave number, $k = 10^7$ m$^{-1}$.
  • The speed of light in vacuum is a constant, $c \approx 3.0 \times 10^8$ m/s.

Calculating Refractive Index

First, calculate the speed of light ($v$) in the medium using the identified $\omega$ and $k$:

$ v = \frac{\omega}{k} $

Substituting the values:

$ v = \frac{2.0 \times 10^{15} \text{ rad/s}}{10^7 \text{ m}^{-1}} = 2.0 \times 10^{8} \text{ m/s} $

Next, calculate the refractive index ($n$) using the calculated speed ($v$) and the speed of light in vacuum ($c$):

$ n = \frac{c}{v} $

Substituting the values of $c$ and $v$:

$ n = \frac{3.0 \times 10^8 \text{ m/s}}{2.0 \times 10^{8} \text{ m/s}} = 1.5 $

Thus, the refractive index of the medium is 1.5.

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Important Questions from Electricity and Magnetism

  1. The electric current in the circuit is given as $i = i_o(t/T)$. The r.m.s current for the period $t = 0$ to $t = T$ is ________.
  2. In the potentiometer, when the cell in the secondary circuit is shunted with $4\text{ }\Omega$ resistance, the balance is obtained at the length $120\text{ cm}$ of wire. Now when the same cell is shunted with $12\text{ }\Omega$ resistance, the balance is shifted to a length of $180\text{ cm}$. The internal resistance of cell is ________$\Omega$
  3. Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \text{ cm}$ length of wire $Q$ is________.

    $(\mu_o = 4\pi \times 10^{-7} \text{ T.m/A})$

  4. For the two cells having same EMF $E$ and internal resistance $r$, the current passing through the external resistor $6\text{ }\Omega$ is same when both the cells are connected either in parallel or in series. The value of internal resistance $r$ is ________$\Omega$.
  5. The magnetic field at the centre of a current carrying circular loop of radius $R$ is $16 \text{ \mu T}$. The magnetic field at a distance $x = \sqrt{3}R$ on its axis from the centre is ________$\text{\mu T}$.
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