 In today's session, we will see the solution to the Maxwell's equations. These are the learning outcome. At the end of this session, student will be able to solve the problem using Maxwell's equations. These are the contents. Before going to solve the problem on Maxwell equation, you should know about the Maxwell equation. So, you can pause video here for a 2 second and recall that what is Maxwell's equation? Define Maxwell's equations. Yes, the Maxwell equation describes the relation between changing electric and the magnetic field. It is the extension of Gauss law, Faraday's law and Ampere's law. We can define different Maxwell's equation for static as well as the time-wearing field by using Gauss law, Faraday's law, Ampere's law, continuity equation, etc. Now let us recall all the Maxwell's equation for static field. Static field is nothing but the field which is not varied with the time. So these are the different laws are defined in first column. For that, we can define the Maxwell's equation in integral form and the derivative form which is also known as a point form. So Maxwell's equation from the Ampere law in integral form as well as the point form it is given like this equation h bar dot dl bar equal to integration j bar dot ds bar which is an integral form whereas del cross h bar equal to j bar is in point form. Similarly, you can recall all other Maxwell's equation already derived in last session which is given by these all equations. By using Faraday's law, this is the equation Gauss law for the electric field. By taking the reference of this Gauss law, we can define the Maxwell's equation in integral as well as the point form. Similarly, the Gauss law for the magnetic field again taking the reference of this magnetic field Gauss law for defining the Maxwell's equation and the last equation is related to the continuity equation. Similarly, you can define all Maxwell's equation for time varying field. So for MPS law, this is the equation h bar dot dl bar equal to integration dow d bar by dow t ds bar meaning of this is what the d bar is the flux density can be differentiated with respect to time and again it is integrated with respect to the surface. So dow d bar by dow t is nothing but you can say that it is the field which is varied with respect to time. Similarly, in the point form you are getting del cross h bar is equal to dow d bar by dow t. Thus by using Faraday's law, you can define the Maxwell's equation in integral form and the point form as shown in this equation. This is the third equation Gauss law for electric field. d bar dot ds bar you are getting equal to 0 whereas del dot d bar is equal to 0. When the del operator is used along with any vector field then it is known as divergence of that field. So that is del dot d bar is the divergence of d bar you are getting that result is equal to 0. Gauss law for the magnetic field it is defined as integration v bar dot ds bar is equal to 0 and del dot v bar equal to 0. So if you know all these equations you can solve the problem on this Maxwell's equation. So let us check one of the example which is given here verify whether or not this following pair of fields satisfy Maxwell's equation in free space. So here the free space is given as the medium and the two equations are given for the electric field intensity as e bar and the magnetic field intensity h bar where e bar is given as a 2 sin x sin t a bar y and h bar is 2 by mu naught cos x cos t a z bar. So if you know these two equations and all Maxwell's equation you have to take the reference of each Maxwell's equation by using the different laws and solve it for the LHS and the RHS that is left hand side of the equation and the RHS that is right hand side of the equation. So again you define the equation for Maxwell's equation for the free space del dot d bar is equal to 0 let us consider this is the equation number 1. Second equation is del dot b bar equal to 0, third equation del cross e bar is equal to minus dou b bar by dou t and the fourth equation del cross h bar equal to dou d bar by dou t. So if you know all these four equations let us consider the given pair of e bar and h bar for each equation. So given e bar is nothing but this one to sin x sin t a y bar and h bar is 2 by mu naught cos x cos t a z bar. Now from this equation if you observe this equation a y bar term is there therefore whatever is the given equation which having the magnitude is in y direction only therefore e bar can be written as e y of which is the function of x and t therefore you can write e y of x and t a y bar as it is. Similarly h bar is the function of h bar equation having the function of x and t only and the magnitude is in the direction of z therefore h bar can be written as h z of x of t a bar z. Now from this equation you can say that since e y and h z are the functions of x and t only therefore the differentiation with respect to y and z are considered as 0. So by considering that as 0 that is the meaning of that is dou by dou y is equal to dou by dou z equal to 0. So wherever you are getting the term partial differentiation with respect to y and partial differentiation with respect to z you have to consider that is equal to 0 because that terms are absent in the given functions e bar and h bar. Now let us consider the first Maxwell equation del dot d bar equal to 0 where d bar is nothing but the electric flux density this is considered as the LHS and the RHS is equal to 0. Now consider LHS first del dot d bar. So the given equation is in terms of e bar. So you know the relation between d bar and e bar as epsilon not e bar. So this equation can be converted into the equation of e bar. So you are getting epsilon not del dot e bar whereas the given equation of e bar which having only the e y term. So epsilon not dou by dou y of e y is equal to 0 why it is equal to 0 because in previous slide we have written that dou by dou y and dou by dou z terms are 0. Therefore you are getting this equation equal to 0 thus it matches with the right hand side. So this is the first Maxwell equation which satisfied here therefore you can say that the given pair e bar and h bar satisfies the first Maxwell equation. Now consider the second Maxwell equation. In second Maxwell equation divergence of b bar is equal to 0 that is del dot b bar equal to 0. b bar is nothing but the magnetic flux density. This is considered as the LHS whereas the RHS side again you getting the result is 0. Now del dot b bar so you know the equation of h bar which is already given. Here also the h bar is dou by dou z of h z is equal to 0. The given pair which having the function with respect to only x and t z term is absent there therefore the differentiation with respect to the z is equal to 0. So here also you are getting the del dot b bar which results into 0 for the given h bar equation. Again from this equation you can say that the given pair satisfies the second Maxwell equation which is nothing but del dot b bar equal to 0. Now consider the third Maxwell equation. Third Maxwell's equation is del cross e bar equal to minus dou b bar by dou t. So del cross e bar is the curl operation whereas dou by dou t of b bar which is negative in the right hand side. So LHS which having the del cross e bar curl operation can be solved by taking the determinant of this e bar equation which is already given. And the RHS which having minus dou b bar by dou t. Now first calculate the del cross e bar how it is represented by using determinant. So the first row is with respect to the unit vector in each axis that is a bar x a bar y and a bar z. Second row is related to the partial differentiation with respect to x, y, z. And the third is nothing but the magnitude of the e bar that is e x, e y and e z. Now solve this determinant. So what you are getting? This is the equation. These two are these two dou by dou y and dou by dou z both are equal to 0. Whereas the e bar equation just having the e y term whereas the x and z are equal to 0. So what you are getting? Del cross e bar is equal to dou by dou x of 2 sin x sin t a z bar which is already given to you. Now if you are taking the differentiation with respect to x, sin x term is there. Sin x differentiation is the cos x. So you are getting the result for del cross e bar as 2 cos x sin t a z bar. Let us consider this is the equation number 5. Now consider the RHS of this equation. It is minus dou by dou t of b bar. So b bar is converted into mu naught h bar because the h bar is known to you. So it can be written as minus mu naught dou by dou t of h bar. So you know the equation of h bar is 2 by mu naught cos x cos t a z bar and minus mu naught as it is and you are taking the differentiation with respect to time now that is dou by dou t. So dou by dou t of cos t is the minus sin t therefore minus minus it becomes as a plus sin and you are getting the result dou by dou t of b bar equal to 2 cos x sin t a z bar. Let us consider this is the equation number 6 and if you are comparing this equation number 5 and 6 which is LHS and RHS respectively from this equation you can say that these two are equal therefore from this equation number 5 and 6 you can say that the given pair satisfies the third Maxwell equation also. Now let us consider the fourth Maxwell equation del cross h bar equal to dou by dou t of d bar. Now this is LHS and this is RHS. Similar to the previous case you have to take the determinant for now the h bar term. So it is given as del cross h bar equal to which having just a h z term. Now let us consider that minus a y bar term is there and dou by dou x of h z term is remaining and others are 0. So if you are taking minus a y bar dou by dou x of h z which having 2 by mu naught cos x cos t. So dou by dou x that is the differentiation with respect to x again the differentiation of cos x is nothing but the minus sin therefore you are getting the plus sin for the result del cross h bar equal to 2 by mu naught sin x sin t a y bar. Now let us consider the RHS of this equation which is dou by dou t of d bar. d bar can be written in terms of e bar as epsilon naught e bar. So epsilon naught dou by dou t of e bar, epsilon naught as it is dou by dou t and you are putting the value of e bar as 2 sin x sin t and a z bar. Now dou by dou t is nothing but the differentiation with respect to time and the time with respect to time the term is the sin t. Sin t differentiation is nothing but the cos t. So by solving this equation you are getting the differentiation with respect to time for d bar as epsilon naught to sin x cos t a y bar. This is the equation number 8. Seventh equation which having the left hand side of the equation and the RHS which is the equation number 8. So if you compare these two equations so these two equations are not same. Therefore you cannot say that this pair which is given which satisfies the Maxwell equation. So remember that you have to check all the Maxwell's equation and although other three equations are matched with LHS and RHS this equation is not matching therefore we say that given pair does not satisfy all Maxwell's equations. Because you can solve the problems on Maxwell's equations. These are the references given for this session. Thank you.