Presentation on theme: "Law of Electromagnetic Induction"— Presentation transcript:
1 Law of Electromagnetic Induction Chapter 6 Electromagnetic InductionLaw of Electromagnetic InductionInductancesEnergy and Force1. Law of Electromagnetic Induction2. Inductances3. Energy in Steady Magnetic Fields4. Magnetic Forces
2 1. Law of Electromagnetic Induction From physics we know that when the magnetic flux through a closed coil is changing, an induced electromotive force e will be generated in the coil, with the relationwhere the positive direction of the electromotive force e and that of the magnetic flux comply with the left hand rule.If the magnetic flux is increased with time, the direction of the induced electromotive force and that of the magnetic flux obey the left hand rule. if the magnetic flux is decreased with time, they will obey the right hand rule.
3 The induced magnetic flux caused by the induced current in the coil always resists the change of the original magnetic flux. The induced magnetic flux is called the reaction magnetic flux, and the induced electromotive force is called the back electromotive force.eIWhen the induced electromotive force is generated in the coil, there is an electric field to push the charges to move in the coil, and this induced electric field is denoted as E.The line integral of the induced electric field intensity around the closed coil is equal to the induced electromotive force in the coil e, i.e.
4 Considering , we haveWhich is called the law of electromagnetic induction, and it shows that when the magnetic field through a closed coil is changing, an induced electric field will be generated in the coil.The law of electromagnetic induction shows that a time-varying magnetic field can produce a time-variable electric field.Based on Stokes’ theorem, from the above equation we haveSince the equation holds for any area S, the integrand must be zero, so that
5 which is called the differential form of law of electromagnetic induction, and it means that the negative time rate of change of the magnetic flux density at a point is equal to the curl of the time-variable electric field intensity at that point.The law of electromagnetic induction is one of basic laws for time-varying electromagnetic fields, and it is also one of Maxwell’s equations.
6 2. InductancesIn a linear medium, the magnetic flux through the closed circuit is also proportional to the current I.The magnetic flux linked with the current I is called the magnetic flux linkage with the current I, and it is denoted as . The ratio of to I is denoted by L, henceIt is called the inductance of the circuit, with the unit henry (H), and the inductance can be also considered as the magnetic flux linkage per unit current.In linear media, the inductance of a circuit depends only on the shape and the sizes, but not on the current.The magnetic flux linkage is different from the magnetic flux, and it is associated with a current.
7 If the magnetic flux is linked with a current N times, then the magnetic flux linkage will be increased by N times. If only a part of the magnetic flux is linked to a current, the magnetic flux linkage must be reduced proportionately.A loop coil with N turns the magnetic flux linkage with the current is = N , and the inductance of the loop coil with N turns isdl1Ozyxdl2l2l1I2I1r2 - r1r2r1Suppose we have two loop currents, the magnetic flux linkage 1 linked with current I1 consists of two parts: one is generated by the magnetic flux caused by current I1 itself, and it is denoted as 11 .Another 12 is produced by the magnetic flux at loop l1 by current I2.
8 Hence, the magnetic flux linkage 1 linked with current I1 is If the surrounding medium is linear, then all the ratios, , ,and are independent of the currents since all the magnetic flux linkages are proportional to the current generating them.Letwhere L11 is called the self-inductance of loop l1, and M12 is called the mutual inductance from loop l2 to loop l1.Similarly, we definewhere L22 is called the self-inductance of loop l2, and M21 is called the mutual inductance from loop l1 to loop l2.
9 Substitute the above parameters L11，L22，M12 ，and M21 into the above equation, we have For linear homogeneous media, we can prove thatSince we can find the mutual inductances between any two loop circuits as follows:Considering ，we have
10 If everywhere, the mutual inductances be zero. If everywhere, the mutual inductances will be maximum.In electronic devices, if we need to increase the magnetic coupling between two coils, the two coils should be placed parallel to each other. If the magnetic coupling needs to be eliminated, they should be perpendicular to each other.The mutual inductance could be positive or negative, while the self-inductance is always positive.
11 The magnetic flux linkage 21 with current I2 by current I1 is Example 1. Calculate the mutual inductance between an infinitely long straight line and a rectangular coil. The line and the coil are at the same plane, and in vacuum.Solution: Select cylindrical coordinate system, and let the infinitely long straight line to be at the z-axis. The magnetic flux density produced by current I1 is thenabdrrD0I1I2zS2The magnetic flux linkage 21 with current I2 by current I1 isIf the flowing direction of the current I2 is clockwise, dS and B1 have the same direction. Then
12 abdrrD0I1I2zS2We haveIf the flowing direction of the current I2 is counter clockwise, then the B1 and dS are opposite, and M21< 0.Example 2: Calculate the inductance per unit length of a coaxial line carrying a direct current.Solution: Assume the radius of the inner conductor of the coaxial line is a, the interior radius of the outer conductor is b, and the exterior radius is c.bcaO
13 The inductance per unit length of a coaxial line is In the coaxial line, we construct a rectangular circuit from a longitudinal section of unit length, left side width a and right width (c b).aIObcraOdrIeThe current in the inner conductor is that on the left side, while the current in the outer conductor is that on the right side.The inductance per unit length of a coaxial line iswhere I is the current in the coaxial line, and is the magnetic flux linkage per unit length with the current I.
14 The magnetic flux linkage with current I is formed by three parts: the first one is in the outer conductor, the second is between the inner and the outer conductors, and the third is in the inner conductor.aIObcrdreSince the thickness of the outer conductor is usually very thin, the magnetic flux linkage in the outer conductor can be neglected.The magnetic flux density Bo between the inner and the outer conductors is given byThe magnetic flux produced by this magnetic field is called outside magnetic flux , then the outside magnetic flux per unit length is
15 The magnetic flux density Bi in the inner conductor is IObcrdreThe outside magnetic flux is linked completely with the current I, and it is equal to the magnetic flux linkage with the current I.The magnetic flux density Bi in the inner conductor isThe magnetic flux produced by this magnetic field is called inside magnetic flux , then the magnetic flux through the area of unit length and width dr is
16 aIObcrdreThis magnetic flux is only linked with the partial current I between 0 and r in the inner conductor, instead of the full current I.Therefore, to the full current I, the magnetic flux linkage formed by this magnetic flux should be scaled down to beFrom this we find the magnetic flux linkage i formed by the magnetic field in the inner conductor that links with the full current I as
17 Then the total magnetic flux linkage with the full current I as Then the total magnetic flux linkage with the full current I as The inductance per unit length of the coaxial line is thereforewhere the first term is called external inductance, and the second term is called internal inductance.when a coaxial line is operating with time-varying electro-magnetic field, the magnetic flux in the inner and the outer conductors can both be neglected.Hence, the inductance per unit length of the coaxial line is equal to the external inductance, given by
18 3. Energy in Steady Magnetic Fields If an impressed source is applied to a circuit, a current will be generated in the circuit. In the process of establishing the current, the reaction magnetic flux in the circuit will resist the increment of the current.In order to overcome the back electromotive force due to the reaction magnetic flux and to maintain the current, the impressed source has to do work.Assume the current is increased very slowly so that radiation loss can be neglected, all energy provided by the impressed source will be stored in the magnetic field around the circuit.Based on the work done by the impressed source, the energy stored in the magnetic field can be calculated.
19 Suppose the current in a loop is gradually increased from zero to the final value I, and the magnetic flux through the loop is also increased from zero to the final value .The back electromotive force isIn order to overcome the back electromotive force, the impressed source has to produce a voltage , given byIf the current in the loop is i(t) at time t, then at this moment the instantaneous power sent to the loop isThe work done by the impressed source in dt is
20 At any moment the relationship between the magnetic flux linkage with a current loop and the current isThe magnetic flux linkage with a current loop is just the magnetic flux through the loop, henceConsidering the inductance L of a loop is independent of the loop current i, we find the work done by the impressed source in dt asWhen the loop current has increased to the final value I, the total work W done by the impressed source is
21 The work establishes the current I in the loop, and the current produces the magnetic field in space.Since the current is increased very slowly, the radiation loss can be neglected, and all work done by the impressed source will be converted into the energy stored in the magnetic field.If the energy of the magnetic field is denoted as Wm , it must be given byorFor some loop circuits, it is very convenience to calculate the inductance by the above equation.Considering , the energy of the magnetic field produced by a loop circuit carrying the current I can be rewritten aswhere is the magnetic flux linkage for the current I.
22 For N current loops, we can let all currents of the loops be increased slowly at the same rate from zero to the final value. Based on energy conservation, the final energy is independent of the process in reaching the final state.Since the magnetic flux linkage of a loop is proportional to the current, the magnetic flux linkage j of the j-th loop isWhen all currents are increased at the same rate, the magnetic flux linkage is also increased at the same rate. Assume the current of the j-th loop at the moment t is , where Ij is the final value, and is the rate coefficient ( ).Then the work done by the impressed source in N loops in dt is
23 When all currents are increased to the final values, the total work W done by the impressed source isThe magnetic energy produced by the N loops carrying the final currents isThat isIf the magnetic flux linkages and the currents of all loops are known, the magnetic energy caused by these loop currents can be determined.The magnetic flux through a circuit can be expressed in terms of the vector magnetic potential A as Hence the magnetic flux linkage with the j-th circuit can also be expressed by the vector magnetic potential A as
24 Then the magnetic energy caused by N loop currents can be expressed in terms of the vector magnetic potential A aswhere A is the composite vector magnetic potential produced by all currents at the j-th circuit.If the currents are distributed continuously in the volume V and, then the right side of the above equation becomes a volume integral. In this case, the magnetic energy can be written asIf the currents are distributed on a surface S, the magnetic energy is given by
25 The magnetic energy density Considering ，we haveUsing the vector identity , the above equation can be rewritten aswhere V is the region in which the current is found.Apparently, if the domain of integration is extended to infinity, the above equation still holds. Let S be the surface of a sphere with infinite radius. Applying divergence theorem, the first integral becomes
26 When the current is distributed in a region of finite volume, the magnetic field intensity at large distance is inversely proportional to the square of the distance, while the vector magnetic potential is inversely proportional to the distance. Hence the surface integral over the surface at infinity will be zero, we haveConsidering ，we havewhere V is the volume occupied by the magnetic field. Obviously, the integrand stands for the density of the magnetic energy.If the magnetic energy density is denoted by small letter wm , then
27 For linear isotropic media, , and the magnetic energy density can be rewritten as Since the magnetic energy is proportional to the square of the magnetic field intensity, the magnetic energy does not obey the superposition principle.Example. Calculate the magnetic energy per unit length in a current-carrying coaxial line. Suppose the steady current is I, the radius of the inner conductor is a, the radius of the outer conductor is b, with thickness neglected, and the region between the conductor is vacuum.
28 Solution: The inductance of a coaxial line per unit length is Hence, the magnetic energy per unit length in the coaxial line asWe can also calculate the magnetic energy in the coaxial line by using magnetic energy density.The magnetic field intensity inside the inner conductor isHence, the magnetic energy inside the inner conductor per unit length is
29 The magnetic field intensity Ho between the inner and the outer conductors is Therefore, the magnetic energy per unit length between the inner and the outer conductors isThe total magnetic energy per unit length in the coaxial line is then , which is the same as before.Since , it is very simple to calculate the inductance of a circuit by using the magnetic energy.
30 The magnetic force F acting on the current element Idl is 4. Magnetic ForcesThe magnetic force F acting on the current element Idl isdl1Ozyxdl2l2l1I2I1r2 - r1r2r1Then the force dF21 of the magnetic field caused by the current I1 on the current element I2dl iswhere the magnetic flux density B1 produced by the current I1 asHence, the force F21 due to the magnetic field B1 on the loop current l2 is
31 In the same way, the force F12 due to the magnetic field B2 caused by the current I2 on the current loop l1 isThe above two equations are called Ampere’s law.Based on Newton’s third law, we have , which can be proved directly from the equations.If the shape of the loop circuit is complex, it is very difficult to evaluate the integral, and a close form expression cannot even be found.In order to calculate the magnetic force, the method of virtual work can be used.Here we directly apply the generalized forces and coordinates to derive the general formulas for obtaining the magnetic force.
32 Under the influence of the generalized force F of the magnetic field caused by the current I1, assume one generalized coordinate of the loop circuit l2 has an increment dl , leading to an increment of the magnetic energy dWm.Then the total work dW done by the impressed source in the two loop circuits should be equal to the sum of the work by the generalized magnetic force and the increment of the magnetic energy so thatTwo cases are to be considered:(a) If the currents I1 and I2 are unchanged, the case of a constant current system results. ThenThe works done by the impressed source for two circuits, respectively, are
33 Then the total work dW done by the impressed source in the two loop circuits is i.e.We find(b) If the magnetic flux linkages with all circuits are unchanged, we have the case of a constant magnetic flux system.Since there is no change in the magnetic flux, no new electromotive force is produced in the two circuits. Hence, the impressed source does zero work, i.e.We find
34 Magnetic forces are used widely, such as in electromagnets, maglev bearings, and maglev vehicles, all of them being based on the action of magnetic forces.Example 1. Calculate the magnetic force between an infinitely long current-carrying wire and a rectangular current-carrying loop. The dimensions and the position of the loop are shown in the figure.Solution: Let it be the constant current system. The magnetic force between the wire and the loop is thereforeabD0I1I2where
35 SinceWe obtainTaking the separation D as the generalized coordinate l, and since the self-inductance L11 and L22 are independent of the separation D, the magnetic force isThe mutual inductance M isWe findwhere the negative sign shows that it is an attractive force.If the direction of one of the currents is opposite, the term involving M becomes negative, leading to a repelling force.
36 Example 2. Find the attractive force of the electromagnet. B0SIlSolution: Since the iron core can be approximately considered as a perfect magnetic conductor, the magnetic field intensity in the iron core is zero, and the magnetic energy is also zero.In this way, the magnetic energy exists only in two gaps. Hence, the total magnetic energy is
37 Considering the magnetic flux in the gaps is , we obtain In view of this, in order to find the attractive force of the electromagnetic magnet, it is easy to consider this case as a constant magnetic flux system. Hence, we havewhere the negative sign shows that it is an attractive force.In addition, we can see that the attractive force is proportional to the cross-sectional area S of the iron core and the square of the magnetic flux density in the gaps.