 Hello everyone, myself Rohini Madhug, working as associate professor in ANTC department at Walshian Institute of Technology, Solapur. Today's topic is Hulse Cut-off Voltage Equation. It is regarding magnetron and magnetron, we are going to derive the equation of cut-off voltage needed and that is derived by HUL, so the name is Hulse Cut-off Voltage Equation. Learning outcomes, at the end of this video you will be able to derive HUL's Cut-off Voltage Equation, you will be able to determine the cut-off magnetic flux density needed to produce oscillations. Now, the question is what is magnetron? Magnetron, it is a self-excited oscillator, it is a cross-fill tube called M-type tube. The direction of electric field that accelerates the electron beam is perpendicular to the axis of magnetic field. Magnetrons are used as radar transmitters and in microwave ovens. Let us revise what is magnetron with its construction and working, let us revise it. Magnetron consists of a cathode at the center, surrounding it are the anode cavities and in this way the electric field will be, electrons will be emitted from cathode moving towards anode and in the perpendicular direction there will be a magnetic field and the oscillations generated inside this cavities are collected using this probe. The same is explained here like when the electric field is applied between anode and cathode without applying the magnetic field, the electron moves a straight path, when small amount of magnetic field is applied which is perpendicular to this electric field, the electron takes the curved path, when the amount of applied magnetic field is increased the electron bends more and if we increase the magnetic field further the electron takes this longer path and this type of electrons will be helpful for oscillations. The electron taking longer path and giving its energy to the RF field and producing oscillations inside the cavities, so magnetron oscillate. So Hull's cutoff voltage equation is regarding this voltage and this magnetic flux density which is needed to produce oscillation. Hull's cutoff voltage equation, what is the Hull's cutoff voltage equation is, this is the voltage VOC provided between anode and cathode, so that electron will just gaze the anode and return towards cathode. The corresponding magnetic flux density is called cutoff magnetic flux density BC. Hull's cutoff voltage equation, so let us start deriving it. Force acting on electron is F is equal to BEV, B is the magnetic flux density, E is the electron charge, V is the velocity, so force can be given by multiplication of these three. In the direction of phi, the force F phi is given as F phi is BEV rho, F phi Y phi because we are interested in the electron which is taking this curvature path, so in the angular path, so in the angular direction the direction is denoted by phi, so it is called as F phi. F phi is force in the angular direction is BEV rho, why V rho? Because rho is the radial direction and the velocity in the radial direction indicates that the electron is taking the path from cathode to anode along the radial direction, so it is V rho. So where B is the applied magnetic flux density is the electron charge, V rho is the velocity in radial direction at a distance rho from center of the cathode cylinder. Torque in phi direction is given as Torque's basic equation is force into distance, when I find torque in phi direction it is rho into F phi, rho into BEV rho, the equation which we have derived just now BEV rho, the same equation is substituted here. So it is rho BEV rho, this is my equation 1. Now rate of change of angular momentum can be given as d by dt of angular velocity into moment of inertia. Angular velocity into moment of inertia is nothing but angular momentum and rate of change is denoted by d by dt. So d by dt into d phi by dt is the angular velocity into moment of inertia is m rho square. So the equation number 2. Rate of change of angular momentum is nothing but the torque in angular direction. These both we have derived, so we can equate these two equation 1 and equation 2. So rho BEV rho is equal to d by dt d phi by dt into m rho square. Now I can integrate this equation with respect to time. So integrating on both side with respect to time, I have written the integral sign here EBV rho into rho dt and when I integrate on this side with respect to time, there is already a differentiation d by dt and I am integrating both cancels each other. So I get d phi by dt into m rho square, C is the constant for integration. So the same equation is written here, only the change is you can see here instead of V rho it is d rho by dt. What does it mean? V rho is the velocity in radial direction. Velocity in radial direction can also be given as rate of change of radial distance or d rho by dt. Now next we can see this dt and this dt get cancelled out. So my equation on the LHS side will be integral EB rho d rho and right hand side will remain same. So I need to integrate this rho d rho or rho with respect to d rho so that my equation will be EB rho square by 2, rho d rho has given me rho square by 2 and right hand side is remain same. This is my equation 3. Now applying boundary condition to this equation 3 at cathode rho is equal to A. A is a cathode radius at cathode rho is A that time the angular velocity d phi by dt is 0. Why it is 0? Because at cathode the electron is just at the verge of emission it has not attained any angular velocity so d phi by dt is 0. Substituting this condition in equation 3 I get C d phi by dt is 0. So I get C is equal to EB instead of rho square it is A square by 2. This is constant C constant of our integration. Now substitute this constant in this equation 3 so I will get this equation. In this you can see this term and the first term are same these two terms are same I can combine these two I can take out the constant as EB by 2 and I can write the further equation like this d phi by dt into M rho square as it is as in the previous equation you can see d phi by dt into M rho square as it is and I am just combining these two. So EB by 2 into rho square minus A square. Now further I want to find d phi by dt so I can take M rho square on this side I will take M below this and rho square I will divide rho by rho square further to this bracket. So I will get EB by 2M 1 minus A square upon rho square because I have divided by rho square here and here. So rho square A square cancels here and A square by rho square this is my equation number 4. According to law of conservation of energy we can say that potential energy is equal to kinetic energy. Potential energy is given as E v0 v0 is the voltage applied between anode and cathode which is equal to one half MV square. This small v is the velocity and capital V is the voltage voltage applied between anode and cathode and this is the velocity attained by the electron. This velocity has two components one is in the radial direction other is in the angular direction. So this v can be written as v rho square plus v phi square. Now I can expand again further this is as it is LHS side as it is RHS is one half M v rho can be written as d rho by dt bracket square but v phi can be written as rho d phi by dt square. Why rho d phi by dt? Because as we have learned in the electromagnetic engineering the cylindrical coordinate system has multipliers as 1 rho 1, 1 is the multiplier with rho, rho is the multiplier with phi and again 1 is the multiplier with z here no question of z so only the multiplier I need to take into consideration with phi. So it is d phi by dt square plus rho d phi by dt square. Now again I apply the boundary condition that rho is equal to b and d rho by dt is 0. How d rho by dt is 0? At rho is equal to b it is at the surface of the anode and radial velocity is reached to 0. At surface of anode radial is reached to 0 so that e v naught is equal to one half M b square d phi by dt bracket square. Substitute this d phi by dt from equation 4. So I can substitute d phi by dt here from equation 4 let us see what is equation 4. This is my equation 4. So this value I can substitute here. So I can substitute one half M b square e b by 2 M 1 minus a square by b square whole bracket square. Instead of rho square I have written b because my boundary condition is rho is equal to b. Here a small mistake is there at anode it should be not at cathode. So I have substituted this and I got this equation this is a substitution part. Now moving further I take e on this side so because I want to find the voltage equation. So instead of writing e v naught I will write v naught and e is taken on this side it is one half M b square as it is a whole bracket square it is e square b square upon 4 M square 1 minus a square by b square whole bracket square. This equation is called as Hull's Cutoff Voltage Equation can be denoted as V OC which is equal to V naught e by 8 M b square b square 1 minus a square upon b square whole bracket square in volts. In this let us see each terminologies a is the cathode radius b is the anode radius e by M this is the ratio will give me some constant because e is the electron charge M is the mass of the electron which are constant so this ratio will give me some value and this value is 1.75 into 10 raise to 11 and this b is applied magnetic flux density it is also called as Cutoff Magnetic Flux Density. Once I found out the Cutoff Voltage Equation I can find out Cutoff Magnetic Flux Density also the corresponding magnetic flux density which is related to this Cutoff Voltage is nothing but the Cutoff Magnetic Flux Density. So I can find out BC by rearranging the terms in this equation BC is 8 V naught M upon e whole bracket rest to 1 half upon b into bracket 1 minus a square upon b square. This is the Cutoff Magnetic Flux Density needed where the electron touches the surface of the anode and returns back V naught capital V naught is the voltage applied between anode and cathode called Cutoff Voltage. We have derived the equation for Cutoff Voltage as well as Cutoff Magnetic Flux Density. These are the references used for preparing this video. Thank you for watching my video.