In an electromagnetic wave in free space the root mean square value of the electric field is V/m. The peak value of the magnetic field is
(b) Given root mean square value of electric field,
We know that, peak value of electric field,
Also, we know that,
where, c= speed of light in vacuum
Out of the following options which one can be used to produce a propagating electromagnetic wave?
(a) A stationary charge (b) A chargeless particle
(c) An accelerating charge (d) A charge moving at constant velocity
(c) An electric charge at rest has electric field in the region around it, but no magnetic field. A moving charge produces both the electric and magnetic fields. If a charge is moving with a constant velocity, the electric and magnetic fields will not change with time, hence no EM wave will be produced. But if the charge is moving with a non-zero acceleration, both the electric and magnetic field will change will space and time, it then produced EM wave. This shows that accelerated charge emits electromagnetic waves.
The energy of the EM wave is of the order of 15 KeV. To which part of the spectrum does it belong?
Given energy of EM waves is of the order of 15keV
i.e. E=hv=h x
=6.624x10-34 x 3x1018/15 x 103 x 1.6x10-19
=1.3248 x 10-29/1.6x10-19
Thus, this spectrum is a part of X-rays.
The condition under which a microwave oven heats up a food item containing water molecules most efficiently in
(a) the frequency of the microwave must match the resonant frequency of the water molecules
(b) the frequency of the microwave has no relation with natural frequency of water molecules
(c) microwave are heat waves, so always produce heating
(d) infrared waves produce heating in a microwave oven
(a) It is electromagnetive wave,therefore the correct choice is (a).
The electric field associated with an electro magnetic wave in vacuum is given by E= 40coswhere E,z and t are in volt/m, metre and second respectively.The value of wave vector k is
(c) (d) 3
Electromagnetic wave equation
Speed of electromagnetic wave v=
Comparing Eqs.(i) and (ii), we get
Here, wave factor k=
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum
is equal to
(a) the speed of light in vacuum
(b) reciprocal of speed of light in vacuum
(c) the ratio of magnetic permeability to the
electric susceptibility of vacuum
so required ratio
Therefore, the ratio of amplitude of magnetic field
to the amplitude of electric field for an electromagnetic
wave propagating in vacuum is equal to reciprocal
of speed of light.