Magnetism Alternating-Current Circuits Alternating Current Generators A coil of area A and N turns rotating with constant angular velocity in a uniform magnetic field produces a sinusoidal emf Alternating current motor Instead of mechanically rotating, we can apply an ac potential difference generated by other ac generator to the coil. This produces an ac current in the coil, and the magnetic field exerts forces on the wires producing a torque that rotates the coil. The magnetic field. Magnetic forces on moving charges. Magnetic forces on a current element Torques on current loops and magnets The Magnetic Field Magnets The Earth is a natural magnet with magnetic poles near the north and south geographic poles. The Magnetic Field Magnets Does an isolated magnetic pole exist? The SI units of magnetic field is the tesla [T] Earth magnetic field is about 10-4 T Powerful laboratories produce fields of 1 -2 T as a maximum 1 Gauss [G] = 10-4 Tesla [T] The Magnetic Field The Magnetic Field. Magnetic force on a moving charge A proton is moving in a region of crossed fields E = 2x105 N/C and B = 3000 G, as shown in the figure. (a) What is the speed of the proton if it is not deflected. (b) If the electric field is disconnected, draw the path of the proton The Magnetic Field. Magnetic force on a current element In the case of a straight segment of length L The Magnetic Field. Torques on Current loops and Magnets B A current-carrying loop experiences no net force in a uniform magnetic field, but it does experience a torque that tends to twist the loop A circular loop of radius 2 cm has 10 turns of wire and carries a current of 3 A. The axis of the loop makes an angle of 30Вє with a magnetic field of 8000 G. Find the magnitude of the torque on the loop. Sources of Magnetic Field The Magnetic Field of Moving Charges The Magnetic Field of Currents. The BiotSavart Law The Magnetic Field Due to a Current loop Sources of Magnetic Field Moving Point Charges are the source of Magnetic Field Sources of Magnetic Field A solenoid A solenoid is a wire tightly wounded into a helix of closely space turns . A solenoid is used to produce a strong, uniform magnetic field in the region surrounded by the loops In this figure, the length is ten times longer than the radius For a long solenoid n = N/L; N number of turns; L : length of solenoid Find the magnetic field at the center of a solenoid of 600 turns, length 20 cm; radius 1.4 cm that carries a current of 4 A Sources of Magnetic Field: Solenoid and magnets Left: Magnetic field lines of a solenoid; right Magnetic field lines of a bar magnet Magnetic Induction Magnetic Flux Induced EMF and FaradayВґs Law LenzВґs Law MAGNETIC INDUCTION In 1830, Michel Faraday in England and Joseph Henry in the USA independently discovered that in a changing magnetic field, a changing magnetic flux through a surface bounded by a stationary loop of wire induces a current in the wire: emf induced and induced current. This process is known as induction. In a static magnetic field, a changing magnetic flux through a surface bounded by a moving loop of wire induces an emf in the wire: motional emf Magnetic flux through a surface bounded by a loop of wire. The SI unit for magnetic flux is weber [Wb] 1Wb= 1 Tв€™m2 Find the magnetic flux through a 40 cm long solenoid with a 2.5 cm radius and 600 turns carrying a current of 7.5 A. MAGNETIC INDUCTION The induced emf is in such direction as to oppose, or tend to oppose, the change that produces it. LenzВґs Law MAGNETIC INDUCTION The coil with many turns of wire gives a large flux for a given current in the circuit. Thus, when the current changes, there is a large emf induced in the coil opposing the change. This self-induced emf is called a back emf Eddy Currents Heat produced by eddy currents constitute a power loss in a transformer. But eddy currents have some practical applications: damping mechanical oscillations, magnetic braking system Inductance Self-inductance The SI unit of inductance is the henry [H] 1 H = 1 Wb/A= 1 T.m2.A-1 Find the self-inductance of a solenoid of length 10 cm, area 5 cm2, and 100 turns Magnetic Energy Alternating Current Generators A coil of area A and N turns rotating with constant angular velocity in a uniform magnetic field produces a sinusoidal emf Alternating current motor Instead of mechanically rotating, we can apply an ac potential difference generated by other ac generator to the coil. This produces an ac current in the coil, and the magnetic field exerts forces on the wires producing a torque that rotaes the coil. Alternating Current Generators A coil of area A and N turns rotating with constant angular velocity in a uniform magnetic field produces a sinusoidal emf пЃ¦ m пЂЅ NBA cos пЃ± and пЃ± пЂЅ пЃ· t пЂ« пЃ¤ пЃ¦ m пЂЅ NBA cos( пЃ· t пЂ« пЃ¤ ) пЃҐ пЂЅ пЂ dпЃ¦m пЂЅ NBA пЃ· sin( пЃ· t пЂ« пЃ¤ ) dt пЃҐ пЂЅпЃҐ peak sin( пЃ· t пЂ« пЃ¤ ) frequency f ; пЃ· пЂЅ 2пЃ° f Alternating Current Circuits: Alternating current in a Resistor Inductors in Alternating Currents Capacitors in Alternating Currents Alternating currents in a Resistor Potential drop across the resistor, VR Current in the resistor I Power dissipated in the resistor, P Average power dissipated in the resistor Paverage V R пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V R , peak cos пЃ· t I пЂЅ P пЂЅ I R пЂЅ I peak cos (пЃ· t ) R 2 2 2 Pav пЂЅ ( I ) av R пЂЅ ( I peak cos (пЃ· t )) av R 2 2 2 V R , peak R cos пЃ· t пѓћ I peak пЂЅ V R , peak R Root-Mean-Square Values Inductors in Alternating Current Circuits The potential drop across the inductor led the current 90Вє (out of phase) VL пЂЅ L dI dt пЂЅ dI dt пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V L , peak cos пЃ· t V L , peak L cos пЃ· t п‚® I пЂЅ V L , peak пЃ·L INDUCTIVE REACTANCE sin пЃ· t пЂЅ I peak пЂЅ V L , peak V L , peak пЃ· L пЃ·L ; cos( пЃ· t пЂ пЃ° 2 I rms пЂЅ V L , rms Instantaneous power delivered by the emf to the inductor is not zero The average power delivered by the emf to the inductor is zero. пЃ· L ) Inductors in Alternating Current Circuits The potential drop across a 40-mH inductor is sinusoidal with a peak potential drop of 120 V. Find the inductive reactance and the peak current when the frequency is (a) 60 Hz, and (b) 2000 Hz I peak пЂЅ V L , peak пЃ· L ; I rms пЂЅ V L , rms пЃ· L INDUCTIVE REACTANCE Instantaneous power delivered by the emf to the inductor is not zero The average power delivered by the emf to the inductor is zero. Capacitors in Alternating Current Circuits The potential drop lags the current by 90Вє VC пЂЅ Q C пЂЅ пЃҐ пЂЅ пЃҐ max cos пЃ· t пЂЅ V C , peak cos пЃ· t пѓћ Q пЂЅ CV C , peak cos пЃ· t I пЂЅ dQ dt пЂЅ пЂ пЃ· CV C , peak sin пЃ· t п‚® I пЂЅ CAPACITIVE REACTANCE I peak пЂЅ V C , peak пѓ¦ 1 пѓ¶ пѓ§ пЃ·C пѓ· пѓЁ пѓё V C , peak пѓ¦ 1 пѓ¶ пѓ§ пЃ·Cпѓ· пѓЁ пѓё cos( пЃ· t пЂ« пЃ° ) 2 ; I rms пЂЅ V C , rms пѓ¦ 1 пѓ¶ пѓ§ пЃ·Cпѓ· пѓЁ пѓё Power delivered by the emf in the capacitor: Instantaneous and average Driven RLC Circuits Series RLC circuit The KirchhoffВґs rules govern the behavior of potential drops and current across the circuit. (a) When any closed-loop is traversed, the algebraic sum of the changes of potential must equal zero (loops rule) (b) At any junction (branch point) in a circuit where the current can be divided, the sum of the currents into the junction must equal the sum of the currents out of the junction (junction rule) Series RLC circuits V app , peak cos пЃ· t пЂЅ L 2 L d Q dt пЂ« dQ dt RпЂ« Q C dI dt пЂ« IR пЂ« Q ; I пЂЅ C пЂЅ V app , peak cos пЃ· t I пЂЅ I peak cos( пЃ· t пЂ пЃ¤ ) dQ dt Power delivered to the series RLC circuit P пЂЅпЃҐ I пЂЅпЃҐ P av пЂЅ 1 пЃҐ 2 P пЂЅ RI peak 2 Pav пЂЅ RI peak cos пЃ· t I peak cos( пЃ· t пЂ ) 2 I peak cos пЃ¤ пЂЅ пЃҐ rms I rms cos пЃ¤ dissipated 2 rms пЃ° пЂЅ V app , rms 2 in the resistor R Z 2 as R / Z пЂЅ cos пЃ¤ and I peak пЂЅ V app , peak / Z P пЂЅ 1 2 V app , peak I peak cos пЃ¤ пЂЅ V app , rms I rms cos пЃ¤ Power factor: cosОґ Series RLC circuits A series RLC circuit with L = 2 H, C =2 ОјF and R=20О© is driven by an ideal generator with a peak emf of 100 V and a frequency of 60 Hz, find (a) the current peak (b) the phase (c) the power factor, (d) the average power delivered; (e) the peak potential drop across each element Phasors Potential drop across a resistor can be represented by a vector VR, which is called a phasor. Then, the potential drop across the resistor IR, is the x component of vector VR, Potential drop across a series RLC circuit V app cos пЃ· t пЂЅ L dI dt пЂ« IR пЂ« Q C In the circuit shown in the figure, the ac generator produces an rms voltage of 115 V when operated at 60 Hz. (a) What is the rms current in the circuit (b) What is the power delivered by the ac generator (c) What is the rms voltage across: points AB; points BC; points CD; points AC; points BD?. A certain electrical device draws 10 A rms and has an average power of 720 W when connected to a 120-V rms 60-Hz power line. (a ) What is the impedance of the device? (b) What series combination of resistance and reactance is this device equivalent to? (c) If the current leads the emf, is the reactance inductive or capacitive? The Transformer Because of the iron core, there is a large magnetic flux through each coil, even when the magnetizing current Im in the primary circuit is very small . The primary circuit consists of an ac generator and a pure inductance (we consider a negligible resistance for the coil). Then the average power dissipated in the primary coil is zero. Why?: The magnetizing current in the primary coil and the voltage drop across the primary coil are out of phase by 90Вє A transformer is a device to raise or lower the voltage in a circuit without an appreciable loss of power. Power losses arise from Joule heating in the small resistances in both coils, or in currents loops (eddy currents) within the iron core. An ideal transformer is that in which these losses do not occur, 100% efficiency. Actual transformers reach 90-95% efficiency Secondary coil open circuit The potential drop across the primary coil is V1 пЂЅ N 1 d пЃ¦ tu rn dt If there is no flux leakage out of the iron core, the flux through each turn is the same for both coils, and then V2 пЂЅ N 2 d пЃ¦ tu rn dt п‚® V2 пЂЅ N2 N1 V1 The Transformer A resistance R, load resistance, in the secondary circuit A current I2 will be in the secondary coil, which is in phase with the potential drop V2 across the resistance. This current sets up and additional flux О¦Вґturn through each turn, which is proportional to N2I2. This flux opposes the original flux sets up by the original magnetizing current Im in the primary. However, the potential drop in the primary is determined by the generator emf According to this, the total flux in the iron core must be the same as when there is no load in the secondary. The primary coil thus draws an additional current I1 to maintain the original flux О¦turn. The flux through each turn produced by this additional current is proportional to N1I1. Since this flux equals вЂ“ О¦Вґturn, the additional current I1 in the primary is related to the current I2 in the secondary by N 1 I1 пЂЅ N 2 I 2 These curents are 180 Вє out of phase and produce counteracting fluxes. Since I2 is in phase with V2, the additional current I1 is in phase with the potential drop across the primary. Then, if there are no losses V1 , rms I 1 , rms пЂЅ V 2 , rms I 2 , rms

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