 This is Dr. Rupali Shalke from Vulture Institute of Technology working as an associate professor. In this video lecture, we are going to discuss on the wave equation in lossy media. In the previous video, we are discussed with the wave equation in a lossless media. After completion of this section, students are able to study or derive the wave equation in lossy media. They can represent the wave equation in phasor form. Also, they can solve the related problems. Now, let us consider a transverse electromagnetic wave traveling in a x direction, having the electric field in a y direction and magnetic field in a z direction and wave is propagating in a x direction. This representation shows that the wave is transverse because the electric field and magnetic field are perpendicular to each other as well as they are perpendicular to the direction of propagation. Now, pause the video, recall what value we are considered for the conductivity for perfect dielectric media and free space. For perfect dielectric media and free space, the sigma value is equal to 0 and there is a variation in the epsilon and mu value. Similarly, the assumption for the lossy media and epsilon will depends on the media. Now, let us derive the wave equation for a lossy media. Now, consider a Maxwell's equation derived from the Faraday's law that is del cross e bar is equal to minus dou b by dou t. In this equation, dou by dou t will substitute as a j omega that is a harmonically varying. And as we know the relation between the b and h that is the value of b will be substituted as a mu h bar. Now, we will solve the equation expanding the curl product. But it will be in a rectangular coordinate a x, a y and a z, dou by dou x, dou by dou z, dou by dou z, e x, e y, e z will be equal to minus j omega mu the h x, h y and h z. As we are assumed, our wave propagating in a x direction that is why the value of dou by dou x will not be equal to 0 and the electric field in a y direction that is why the value of e y will also be not equal to 0 and magnetic field in a z direction that is why h z will also be not equal to 0. While other terms are we are considering it as a 0. Now, after solving this determinant, the equation will reduce to dou by dou e y by dou x will be equal to minus j omega mu h z. We will mark it as an equation 1. Similarly, let us consider a Maxwell's equation derived from the Ampere's law which is nothing but a del cross h bar is equal to sigma e bar plus dou d by dou t which is equal to now the dou by dou t will substitute j omega and the relationship between the e and d is d bar is nothing but a epsilon e bar. We will substitute the value of d bar is equal to if you see this equations this is relating between the magnetic field and the electric field. Similarly, we will expand del cross h bar and by same assumption we will substitute the value here in this case dou h z is not equal to 0 now and dou sorry e y will also be not equal to 0 while other terms we will substitute it as a 0 and by considering the common e bar we will substitute the consider equation sigma plus j omega epsilon into other terms. Now solving this equation we get minus dou h z by dou x a y bar is equal to sigma plus j omega epsilon e y a y bar a y will be common that is why the equation will reduce to dou h z by dou x will be equal to minus sigma plus j omega epsilon e y. We will mark this equation as equation 2 we will this use this equations and solve the next problem considering the equation 2 differentiate this equation 2 with respect to x then it will be a second order derivation dou square h z by dou x square will be equal to this term as it is the equation dou by dou e y by dou x we will mark it as an equation 3. Now, substituting the value of dou e y by dou x from the equation 1 we will get the equation dou square h z by dou x square is equal to j omega mu which is a value of dou by dou x h here we will substitute this if you see this equation at the both the side LHS and RHS we are having the value h z that is why we are calling this as a wave equation for the lossy media in the magnetic field or h z term. Similarly, let us consider equation 1 which we are already derived differentiating different equation 1 with respect to x this will be a second order del square e y by dou x square is will be equal to minus j omega mu here dou h z by dou x we will mark it as an equation 4 then substituting the value of dou h z by dou x from equation 2 this value now we will substitute in the equation 4 then the equation 4 will be in this way which is nothing but a second order equation in a terms of electric field. This was in a magnetic term and this is in a electric term this is a wave equation for the lossy media in e y term these both equations are also called as a emboldt equations in terms of h z this term we can represent in a phasor form by substituting the dou square del x by the del operator as it is a second order it will be a del square h z and this term is equal to the propagation constant gamma square h z. If you see this is a phasor equation these both equations are in terms of h z term that is in a magnetic field. Similarly, we can write the wave equation for a lossy media in terms of electric field pause the video think for a while and solve this try to solve this problem. In this case a magnetic field is given in a form of equation which is in a wave direction and the value for the mu r epsilon r and sigma r given and determine the value for the propagation constant that is gamma attenuation constant and beta. As this value it is the equations are for the lossy media the sigma is having some value this is not equal to be 0 it is having some value. Therefore, the it is seems that the when the waves are travelling into the free space in a lossless media then it is a there is no change in the amplitude while when the waves are travelling in the lossy media then there is a losses takes place or attenuation takes place in the amplitude there is a change in the amplitude variations while other term that is a velocity wavelength and beta will remain same. These are the few references. Thank you.