If Electric field intensity of a uniform plane electromagnetic wave is given as $$E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}$$. Then magnetic intensity 'H' of this wave in Am$$-$$1 will be :
[Given : Speed of light in vacuum $$c = 3 \times {10^8}$$ ms$$-$$1, Permeability of vacuum $${\mu _0} = 4\pi \times {10^{ - 7}}$$ NA$$-$$2]
In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $${\lambda \over {100}}$$, where $$\lambda$$ is the wavelength of the wave in free space. The phenomenon, which happens there will be :
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?
The I-V characteristics of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is :