 and today we will solve the numerical on magnetic circuit. These are the learning outcomes of this video lecture. At the end of this session students will be able to first explain the parameters of magnetic circuit and second students will be able to solve numerical on magnetic circuit. These are the contents of this video lecture. Now before moving towards to the example of the magnetic circuit, you have to recall terms related to magnetic circuit and write down the formulae of magnetic circuit. Pause this video and you have to recall the terms related to magnetic circuit. Now we will see the terms related to magnetic circuit one by one. So first one is magnetic flux. Magnetic flux is represented by phi. Magnetic field lines produced by the magnetic source is called magnetic flux. So flux is equal to B into A. So B is nothing but it's the flux density and A is the area. Then second magnetic field strength. This is denoted by capital H and it is magneto motive force per length. Then third parameter MMF. So MMF is nothing but it's a magneto motive force. Magneto motive force is drives the flux through the magnetic circuit. Same as EMF in electric circuit. So here MMF is equal to N into I and it's nothing but it is a number of turns and I is the current. Then the fourth one is reluctance. So reluctance is nothing but it is the property of the magnetic circuit which opposes the creation of flux in it. It is denoted by S. S is equal to L upon mu A. So L is the length, mu is the permeability and the A is the area. Then next parameter that is magnetic flux density. So magnetic flux density it is represented by B. B is equal to mu into H. So mu is the permeability, absolute permeability and H is nothing but it's the magnetic field strength. And from this equation also we can write down B is equal to phi upon A, flux upon area. So according to requirement of the example we will take the equation. Then next parameter that is the permeability. So permeability is nothing but ability of the material to carry the flux lines. So absolute there are the two types means absolute permeability is represented by mu. Then relative permeability is represented by mu R and in free space we have to represent with the help of mu 0. And in free space its mu 0 is nothing but 4 pi into 10 raise to minus 7. It is constant term. Now we will see the example based on the magnetic circuit. So first we will see the first example. The simple magnetic circuit shown has a cross sectional area of 50 centimeter square. Then mean length of 2 meter relative permeability that is mu R is given 100. Then the coil has 250 turns means capital N is given 250 and the flux produced is 100 microveber. So you can see in this figure then what we have to find. So first one reluctance of the magnetic circuit then second current flowing through the coil. Now we will solve the example. So this is given. So first we have to find out reluctance of the magnetic circuit. All of you know that S is equal to L upon mu A and mu is equal to mu 0 into mu R. So mu 0 is the constant term that is 4 pi into 10 raise to minus 7 mm and mu R is given that is 100. So from these two values you have to put into the equation and we will get mu is equal to 4 pi into 10 raise to 5 meter. Then area is given and L is also we know that L is 2 meter. So all the values you have to put into the equation. Calculate this equation and we will get S is equal to 3.18 into 10 raise to 6 ampere per vapor. So this is the reluctance of the magnetic circuit. Then we will find out actually we want to find out current but for current we will do some calculation for that. So H is equal to what mmf upon L. mmf is what magneto motive force. So mmf is equal to what N into I that is number of turns into current divided by L. From this we can calculate the current. So I is equal to H into L divided by N. So again from this we can write down H is equal to B upon mu. So this is equation 1 and we know that B is equal to what 5 upon area. So this is equation 2. So with the help of these two equations here we can write H is equal to 5 upon A into mu. So these terms are given. So you have to put into the equation and we will get H is equal to 159.15 ampere per meter. So this is the magnetic field string. Now we want to calculate what current. Our aim is what we want to calculate the current. We will get all the terms we have to put into the equation and we will get the current. So I is equal to 1.27 ampere. So in this way we have to calculate the reluctance as well as the current of the magnetic circuit. Now we will see example 2. Here you can see in this figure iron ring of mean circumference 50 centimeter. Then air gap is given 0.1 centimeter length. Then windings are 300 turns. Then mu R is also given and current is 1 ampere flows through the coil. Find the flux density in the air gap. So how to calculate the flux density in the air gap we will see. So first we have to note down which terms are given. So first upon you have to remember B is the flux density on the iron ring and its value also would be the same. So B would be the same in the iron ring also and in the air gap also. Okay. So first you have to consider the iron ring and we have to calculate the H1. So H1 is equal to B upon mu R into mu 0. So put the values into that and we will get H is equal to 1,989.44 B. This is the equation 1. B is unknown. Then L1 is given. L1 is equal to 50 centimeter. So you have to convert into the meter 0.5 meter. After that we will calculate the MMF. Okay. So MMF across the iron ring. So all these terms related to iron ring. So MMF is equal to what? H1 into L1. So from equation 1 and 2 put the values into that and we will get F1 is equal to 995 B. So now you have to consider the air gap. So before that we have to calculate it for the ring. Now we have to calculate for the air gap. So for air gap we will calculate H2. H2 is equal to B upon mu 0. Here mu are not considered. So put the values into that and you have to find out. H2 is what? 7.96 into 10 raise to 5 B. This is equation 1. Then L2 for the air gap is what? 0.1 centimeter. So it is equal to 10 raise to minus 3 meter. It's equation 2. Then after this we have to calculate the MMF across the air gap. So the equation will be F2 is equal to H2 into L2. So put the values from equation 1 and 2 and then we will get F2 is equal to what? 7.96 B. Now we want to calculate the total MMF. So total MMF is equal to what? F is equal to F1 plus F2. So you have to add the terms F1 and F2 and we will get 1,791 B. So this is the total EMF. This is equation 3. Total MMF, how to calculate? MMF is equal to what? Number of turns into the current. Okay. So number of turns are given that is 300 into current is 1 ampere. So MMF is equal to 300. So this is equation 4. So from equation 3 and 4 we will get B is equal to 300 upon 1795. So here flux density is equal to 0.1675 Weber per meter square. So in this way we have to calculate the flux density. These are the references of this video lecture. Thank you.