Chap 21. Electromagnetic Induction


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1 Chap 21. Electromagnetic Induction Sec. 1  Magnetic field Magnetic fields are produced by electric currents: They can be macroscopic currents in wires. They can be microscopic currents ex: with electrons in atomic orbits. Magnetic field sources are essentially dipolar in nature, having a north and south magnetic pole Magnetic field lines are continuous and originate from the north pole and terminate at the south pole. The interaction of magnetic field with charge leads to many practical applications. Bar Magnet F = q v B Magnetic field sources are essentially dipolar in nature, having a north and south magnetic pole Magnetic field lines are closed lines and permanent magnets can be made from ferromagnetic materials. Magnetic monopoles do not exist. Compare with a electric charge above
2 Lorentz Force (F) Both electric field and magnetic field can be defined in terms of force on charge in the Lorentz force law. F E = q E FB = q v B F = qvbsinθ (1) The SI unit for magnetic field (B)is the Tesla and it is a vector quantity A smaller magnetic field unit is the Gauss (1 Tesla = 10,000 Gauss). The SI unit for Electric field (E) is the Newton/Coulomb and it is a vector quantity The direction of the electric force is in direction of E if the charge q is positive. If q is negative then direction of FE is reverse of E. The direction of the magnetic part of the force is given by the right hand rule. 2 The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field. The force is maximum when v is perpendicular to B
3 3
4 Force on a current carrying wire 4
5 Ampere s Law 5 The magnetic field in space around an electric current is proportional to the electric current which serves as its source, just as the electric field in space is proportional to the charge which serves as its source. Ampere s Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop. Application We can use Ampere s Law to obtain the following The magnetic field inside a solenoid. The magnetic field around a current carring wire. The magnetic field inside a toriodal coil. The magnetic field inside a conductor.
6 The magnetic field around a current carring wire 6 Use Ampere s Law: Find the component of B tangent to the circle of radius r and sum its value around a closed path. [B is a constant at a circle of radius r (symmetry of the problem)]. B dl = µ o I B 2πr = µ o I B = µ oi 2πr RIGHT HAND RULE 2: If the thumb gives the direction of the current then the curled fingers give the direction of the magnetic field.
7 7 Solenoid: The magnetic field inside a loop of N turns Use Ampere s Law: In the path shown, Find the component of B tangent to the path of the rectangle and sum its value around a closed path.if I is the current in each loop then the total current enclosed in the area is B dl = µ o I enc I enc = In = I N L l B l = µ o IN l L B = µ oin L RIGHT HAND RULE 2: If the curled fingers gives the direction of the current then the thumb give the direction of the magnetic field.
8 Faradays Law: Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be induced in the coil. No matter how the change is produced, the voltage will be generated. The change could be produced by changing the magnetic field (B) strength, moving a magnet toward or away from the coil, moving the coil into or out of the magnetic field, rotating the coil relative to the magnet etc: Changing the area (A) of the coil so that the flux varies. Faraday s law is a fundamental relationship which comes from Maxwell s equations. The induced emf is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. The emf induced in the coil is given as EMF (V ) = dφ dt = d(b A) dt Remember to take the perpendicular component of B to the surface area you consider. The parallel component of B does not induce emf. 8
9 9
10 Voltage generated in a moving coil 10 The force acting on the charges due to the change in the magnetic flux if the wire sweeps an area of A = lx is EMF (V ) = d(b A) dt = Bl d(x) sinθ = Blvsinθ dt
11 AC Generator The turning of a coil in a magnetic field produces motional emfs in both sides of the coil which add. Since the component of the velocity perpendicular to the magnetic field changes sinusoidally with the rotation, the generated voltage is sinusoidal or AC. This process can be described in terms of Faraday s law when you see that the rotation of the coil continually changes the magnetic flux through the coil and therefore generates a voltage. The emf induced for N number of turns is 11 EMF (V ) = 2NBlvsinθ From rotational motion we can write v = ωr = ωh/2 and θ = ωt. EMF (V ) = NBlhω sin(ωt) = NBAω sin(ωt) The maximum induced on peak voltage is when sin(ωt) = 1 (V peak ) = NBlhωsin(ωt) = NBAω
12 Inductance 12 Inductance is typified by the behavior of a coil of wire in resisting any change of electric current through the coil. Arising from Faraday s law, the inductance L may be defined in terms of the emf generated to oppose a given change in current: EMF (V ) = L di dt The unit of inductance (L) is Henry. The inductance of a solenoid of N turns can be given as EMF (V ) = L di dt = N d(b A) dt L di dt = N d(µ onia/l) dt This gives L = µ on 2 A l
13 Transformers A transformer makes use of Faraday s law and the ferromagnetic properties of an iron core to efficiently raise or lower AC voltages. Remember very important... it of course cannot increase power so that if the voltage is raised, the current is proportionally lowered and vice versa. 13 If V s and I s is the voltage and current of the secondary with N s turns and V p and I p is the voltage and current of the primary with N p turns, then N s = V s = I s N p V p I p The power in the primary and secondary is equal P p = P s = I s V s = I p V p In Step up transformer V s > V p and in a step down transformer V p > V s
14 Energy stored in a inductor and a magnetic field 14 The energy stored in an inductor is given as U = P.dt = IV dt = I(L di dt )dt = LIdI = 1 2 LI2 Use the inductance and current derived for a soleniod and it will lead us to the energy stored in a magnetic field. U = 1 2 µ o N 2 A ( Bl l µ o N )2 = 1 B 2 Al 2 µ o The energy density is the energy per unit volume V = Al u = U V = 1 2 B 2 µ o
15 AC circuits  AC response of Resistors 15 For ordinary currents and frequencies the behavior of a resistor is that of a dissipative element which converts electrical energy into heat. It is independent of the direction of current flow and independent of the frequency. So we say that the AC impedance of a resistor is the same as its DC resistance. That assumes, however, that you are using the rms or effective values for the current and voltage in the AC case. The phasor diagram denotes the phase difference of the current and the voltage in the ciruit. The phasor diagram for the same is given below Resistance, Impedance for a resistance is X r = R
16 AC circuits  AC response of Capacitors 16 You know that the voltage across a capacitor lags the current because the current must flow to build up the charge, and the voltage is proportional to that charge which is built up on the capacitor plates. Impedance for a capacitor is X c = 1 ωc For DC currents ω = 0 and hence the Impedance of the capacitor for DC currents is X c = 1 ωc = There is no current in the circuit after the capacitor is charged. Therefore if there in an RC circuit hooked to a DC source, all the voltage is dropped across the capacitor. Voltage across the resistor is zero since the current is zero V R = IR = 0R = 0
17 AC circuits  AC response of Inductors 17 You know that the voltage across an inductor leads the current because the Lenz law behavior resists the buildup of the current, and it takes a finite time for an imposed voltage to force the buildup of current to its maximum. Impedance for a inductor is X L = ωl For DC currents ω = 0 and hence the Impedance of the inductor for DC currents is X L = ωl = 0 Therefore if there in an RL circuit hooked to a DC source, all the voltage is dropped across the resistor. Voltage across the resistor is E = V R = IR. Voltage across the inductor is V L = IX L = I0 = 0.
18 AC circuits  AC response of LRC Circuits 18
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