The provided magnetic field of the electromagnetic wave is given by the equation:
$B = 0.3\cos(ky-10^8t)i$ tesla
This equation represents a plane wave propagating along the positive y-axis. We can compare this to the general form of a magnetic field wave:
$B = B_0 \cos(k y - \omega t)\hat{i}$
From the given equation, we can identify the following parameters:
The electromagnetic wave is propagating through a dielectric medium with a refractive index ($n$) of 2. The speed of the wave ($v$) in this medium is related to the speed of light in vacuum ($c$) and the refractive index ($n$) by the formula:
$v = \frac{c}{n}$
We know the standard value for the speed of light in vacuum is $c \approx 3 \times 10^8$ m/s.
The relationship between angular frequency ($\omega$), wave speed ($v$), and wave number ($k$) is:
$\omega = v k$
The wavelength ($\lambda$) is related to the wave number ($k$) by:
$k = \frac{2\pi}{\lambda}$
Substituting the expression for $k$ into the angular frequency equation:
$\omega = v \left(\frac{2\pi}{\lambda}\right)$
Now, we can rearrange this formula to solve for the wavelength ($\lambda$):
$\lambda = \frac{2\pi v}{\omega}$
Substitute the relation $v = c/n$ into the wavelength formula:
$\lambda = \frac{2\pi (c/n)}{\omega} = \frac{2\pi c}{n\omega}$
Now, we substitute the known values into the derived formula:
Calculation:
$\lambda = \frac{2\pi \times (3 \times 10^8 \text{ m/s})}{2 \times (10^8 \text{ rad/s})}$
$\lambda = \frac{2\pi \times 3 \times 10^8}{2 \times 10^8}$ m
$\lambda = 3\pi$ m
Therefore, the wavelength of the electromagnetic wave in the dielectric medium is $3\pi$ meters.
For a given system of resistors having resistances R, 2R, R$_0$ and 2R (shown in the figure), what will be the value of resistance of the resistor R$_0$, when there is NO current in the galvanometer G?
