Electromagnetic Induction
Electromagnetic inductionThe phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction.
Total number of magnetic field lines passing normally through a given area is called the magnetic flux.Magnetic Fluxπœ™B=Bβ†’.Aβ†’=BAcosπœƒUnit of magnetic flux= T.m2=WeberGauss’s law for magnetism is: The net magnetic flux through any closed surface is zero.
Magnetic Potential energyU=-mBsinπœƒ
Faraday’s law of electromagnetic induction : The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. E=-dπœ™Bdt
If there is a closely wound coil of N turns, then change of flux associated with each turn :E=-Ndπœ™Bdt
magnetic dipole moment (m) associated with a current loop=m=NIA
Motional Electromotive ForceE=BlvI=Er=BlvrForce=F=IlB=B2l2vrPower=P=Fv=B2l2v2ralternatevely,Power=P=I2r=B2l2v2r
Law of electromagnetic inductionFaraday's Law:1. Whenever there is change in the magnetic flux associated with a coil an emf is induced in the coil2. The magnitude of the induced emf is directly proportional to the rate of change of the magnetic flux thru the coil/ curcuit.E=-dπœ™Bdt
Self InductanceIt is also possible that emf is produced in a single isolated coil due to change of flux thru the coil by means of varying current thru the same coil. This is called self induction. Self Inductance = LNπœ™B=LI⟹L=Nπœ™BI E=-dNπœ™Bdt E=-LdIdtSI unit of self inductance (L) = VoltAmpere/second=H(Henry) Work needs to be done against the self induced emf as it opposes any change in current in circuit. This work done is stored as magnetic potential energy.W=12LI2
1. The self-inductance of the coil depends on its geometry and on the permeability of the medium.2. Inductance is intrinsic material properties.3. Inductance is scalar quantity4. SI unit of inductance : henry
Self Inductance of a long solenoid Nπœ™B=(nl)(πœ‡0nI)(A)Nπœ™B=πœ‡0n2AlIL=Nπœ™BI=πœ‡0n2AlL=πœ‡rπœ‡0n2Al
When flux thru the coil changes then each turn contributes to the induced emf. Mutual InductanceWhen changing current in coil produces emf in another coil due to flux change then it is called mutual inductance.Mutualinductance=M=πœ‡0n1n2πœ‹r12l wherer1=radiusofinnersolenoidn1=numberofturnsperunitlengthofinnersolenoidl=lengthofsolenoid N1πœ™1=MI2d(N1πœ™2)dt=d(MI2)dtℇ1=-dN1πœ™1dtℇ1=-MdI2dt
Lenz's Law : The polarity of induced emf is such that it tends to produce a currentwhich opposes the change in magnetic flux that produced it.
Displacement CurrentDisplacement current is different from the conduction current. It is produced when electric current or electric flux varies with time.
TransformerPrinciple: Whenever the magnetic flux linked with a coil changes an emf is induced in the neighbouring coil. It consists of :a. Primary Coil : input coil.b. Secondary Coil : output coil ep:Inducedemfinprimarycoiles:InducedemfinsecondarycoilNp:NumberofturnsinprimarycoilNs:Numberofturnsinsecondarycoil esep=NsNp For ideal transformerInput Power = Output Power epip=esis esep=ipis esep=NsNp=ipis Step Up Transformer: If Ns>Np then it is called step up transformer.Step Down Transformer: If Ns<Np then it is called step down transformer.
Questions and Answer
Example 6.11 Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a 100 turn coil of area 0.1m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?Answer 6.11 ℇ=-Ndπœ™Bdtπœ™B=BAcosπœƒ=BAcosπœ”tℇ=-Nd(BAcosπœ”t)dt=BANπœ”sinπœ”tπœ”=πœ‹1rad/secℇmax=BANπœ”=0.01β¨―0.1β¨―100β¨―πœ‹β„‡max=0.314volt
Q.6.3.A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? A.6.3 Magnetic field inside long solenoidB=πœ‡0nIwhere n=number of turns per unit length n=15 turns per cm=1500 per meterdIdt=4-20.1=20A/sec B=4πœ‹Γ—10-7Γ—1500Γ—IB=6πœ‹Γ—10-4Iπœ™B=6πœ‹Γ—10-4Γ—2Γ—10-4Γ—Iπœ™B=12πœ‹Γ—10-8I ℇ=-dπœ™Bdt=-12πœ‹Γ—10-8dIdtℇ=-12πœ‹Γ—10-8Γ—20Magnitudeofemf,ℇ=|-7.536Γ—10-6|ℇ=7.536Γ—10-6V
Q.6.10A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5β¨―10-5T and the dip angle is 30Β°.A.6.10Speedofplane=1800km/hr=500m/secspanofthewing=25mangleofdip=30Β°B=5β¨―10-5sin30Β°=2.5β¨―10-5Tℇ=Blv=2.5β¨―10-5β¨―0.25β¨―500ℇ=3.125β¨―10-3Volt
Q.6.12A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8cmsec-1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3Tcm-1 along the negative x-direction (that is it increases by 10-3Tcm-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10-3Tsec-1 . Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΞ©.A.6.12HereBischangingwithtime(t)andwithx.L=sideofthesquaredBdx=-10-3Tcm-1=-0.1Tm-1dBdx=-10-3Tsec-1dxdt=8cmsec-1=0.08msec-1πœ™B=B.AEmfduetochangingofmagneticfieldwithtimeℇ1=-d(BA)dt=-AdBdt=-(0.12β¨―0.12)β¨―(-10-3)=0.0144β¨―10-3=1.44β¨―10-5Volt
Q.6.14 (NCERT)Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΞ©. Assume the field to be uniform.(a) Suppose K is open and the rod is moved with a speed of 12cms-1 in the direction shown. Give the polarity and magnitude of the induced emf.(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.(d) What is the retarding force on the rod when K is closed?(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12cms-1) when K is closed?How much power is required when K is open?(f ) How much power is dissipated as heat in the closed circuit? What is the source of this power?(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
A.6.14(a)E=Blv=0.5β¨―0.15β¨―0.12volt=0.009V=9mV.HererodPQisactinglikesourceofEMF.PwillbepositiveandQnegative. (b) K is open: Excess charge is built up when K is open.K is closed: When key is closed current starts flowing. Excess charge is maintained by flow of current. (c)With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ. There is presence of excess charge at P and Q. This sets up an electric field in direction PQ. Now as rod PD is moving so there is also a magnetic force acting in direction QP FB=qVB. Magnetic force on the electrons is balanced by force on them due to force by the electric field. Hence there is no net force on the rod. (d)Rod length = l = 15 cm = 0.15 m Magneticfield=0.5TVelocityoftherod=12cms-1E=Blv=0.5β¨―0.15β¨―0.12volt=0.009VI=VR=0.0099β¨―10-3=1AMagneticforceoncurrentcarrtyingconductor=F=Ilβ†’β¨―Bβ†’βŸΉF=1β¨―0.15β¨―0.5=0.075N (e)Power=P=Fβ¨―v=0.075β¨―0.12=0.009Watt (f)Power dissipitated as heat=P P=VI=I2R=12β¨―0.009=0.009Watt (g) If magnetic field id parallel to the rod then motion of the rod des not cut the magnetic field lines. So there will be no EMF induced.
Q.6.15An air-cored solenoid with length 30 cm, area of cross-section 25cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid. A.6.15Magnetic field inside long solenoid B=πœ‡0nIwhere n=number of turns per unit length Given n=5000.3=50003permeter A=25cm2=25Γ—10-4m2 dIdt=2.5-010-3=2.5Γ—103A/sec B=4πœ‹Γ—10-7Γ—50003Γ—I B=23πœ‹Γ—10-3I πœ™B=23πœ‹Γ—10-3Γ—25Γ—10-4Γ—I πœ™B=53πœ‹Γ—10-6Γ—I ℇ=-Ndπœ™Bdt=-500Γ—53πœ‹Γ—10-6dIdt ℇ=-500Γ—53πœ‹Γ—10-6Γ—2.5Γ—103Magnitudeofemf,ℇ=|-6.541|ℇ=6.541V