Chapter 26 DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram.

Chapter 26 DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram below shows a possible amplifier circuit for each stereo channel. Although the large triangle is an amplifier chip containing transistors (discussed in Chapter 40), the other circuit elements are ones we have met, resistors and capacitors, and we discuss them in circuits in this Chapter. We also discuss voltmeters and ammeters, and how they are built and used to make measurements.

26-5 Circuits Containing Resistor and Capacitor (RC Circuits)
When the switch is closed, the capacitor will begin to charge. As it does, the voltage across it increases, and the current through the resistor decreases. Figure After the switch S closes in the RC circuit shown in (a), the voltage across the capacitor increases with time as shown in (b), and the current through the resistor decreases with time as shown in (c).

The voltage across the capacitor is VC = Q/C: The quantity RC that appears in the exponent is called the time constant of the circuit:

The current at any time t can be found by differentiating the charge:

Example 26-11: RC circuit, with emf. The capacitance in the circuit shown is C = 0.30 μF, the total resistance is 20 kΩ, and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value, (e) the maximum current, and (f) the charge Q when the current I is 0.20 its maximum value. Solution: a. The time constant is RC = 6.0 x 10-3 s. b. The maximum charge is the emf x C = 3.6 μC. c. Set Q(t) = 0.99 Qmax and solve for t: t = 28 ms. d. When Q = 1.8 μC, I = 300 μA. e. The maximum current is the emf/R = 600 μA. f. When I = 120 μA, Q = 2.9 μC.

Example 26-11: RC circuit, with emf.

If an isolated charged capacitor is connected across a resistor, it discharges: Figure For the RC circuit shown in (a), the voltage VC across the capacitor decreases with time, as shown in (b), after the switch S is closed at t = 0. The charge on the capacitor follows the same curve since VC α Q.

Once again, the voltage and current as a function of time can be found from the charge: and

Example 26-12: Discharging RC circuit. In the RC circuit shown, the battery has fully charged the capacitor, so Q0 = CE. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 μF. The current I is observed to decrease to 0.50 of its initial value in 40 μs. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R? (c) What is Q at t = 60 μs? Solution: a. At t = 0, Q = CE = 20.4 μC. b. At t = 40 μs, I = 0.5 I0. Substituting in the exponential decay equation gives R = 57 Ω. c. Q = 7.3 μC.

Conceptual Example 26-13: Bulb in RC circuit. In the circuit shown, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later. Solution: When the switch is closed, the current is large and the bulb is bright. As the capacitor charges, the bulb dims; once the capacitor is fully charged the bulb is dark.

26-7 Ammeters and Voltmeters
An ammeter measures current; a voltmeter measures voltage. Both are based on galvanometers, unless they are digital. The current in a circuit passes through the ammeter; the ammeter should have low resistance so as not to affect the current. Figure An ammeter is a galvanometer in parallel with a (shunt) resistor with low resistance, Rsh.

A voltmeter should not affect the voltage across the circuit element it is measuring; therefore its resistance should be very large. Figure A voltmeter is a galvanometer in series with a resistor with high resistance, Rser.

An ohmmeter measures resistance; it requires a battery to provide a current. Figure An ohmmeter.

Summary: An ammeter must be in series with the current it is to measure; a voltmeter must be in parallel with the voltage it is to measure.

Example 26-17: Voltage reading vs. true voltage. Suppose you are testing an electronic circuit which has two resistors, R1 and R2, each 15 kΩ, connected in series as shown in part (a) of the figure. The battery maintains 8.0 V across them and has negligible internal resistance. A voltmeter whose sensitivity is 10,000 Ω/V is put on the 5.0-V scale. What voltage does the meter read when connected across R1, part (b) of the figure, and what error is caused by the finite resistance of the meter? Solution: On the 5-V scale, the meter has a resistance of 50,000 Ω. The equivalent resistance of the meter and R1 in parallel is 11.5 kΩ. The total resistance of the circuit is 26.5 kΩ, so the current is 0.30 mA; the voltage across R1 (and therefore also across the voltmeter) is 3.5 V. Without the voltmeter, the voltage across R1 is half the battery voltage, or 4.0 V; this is more than a 10% error.

Summary of Chapter 26 A source of emf transforms energy from some other form to electrical energy. A battery is a source of emf in parallel with an internal resistance. Resistors in series:

Summary of Chapter 26 Resistors in parallel: Kirchhoff’s rules:
Sum of currents entering a junction equals sum of currents leaving it. Total potential difference around closed loop is zero.

Summary of Chapter 26 RC circuit has a characteristic time constant:
To avoid shocks, don’t allow your body to become part of a complete circuit. Ammeter: measures current. Voltmeter: measures voltage.

Chapter 27 Magnetism Chapter 27 opener. Magnets produce magnetic fields, but so do electric currents. An electric current flowing in this straight wire produces a magnetic field which causes the tiny pieces of iron (iron “filings”) to align in the field. We shall see in this Chapter how magnetic field is defined, and that the magnetic field direction is along the iron filings. The magnetic field lines due to the electric current in this long wire are in the shape of circles around the wire. We also discuss how magnetic fields exert forces on electric currents and on charged particles, as well as useful applications of the interaction between magnetic fields and electric currents and moving electric charges.

Units of Chapter 27 Magnets and Magnetic Fields
Electric Currents Produce Magnetic Fields Force on an Electric Current in a Magnetic Field; Definition of B Force on an Electric Charge Moving in a Magnetic Field Torque on a Current Loop; Magnetic Dipole Moment

Units of Chapter 27 Applications: Motors, Loudspeakers, Galvanometers
Discovery and Properties of the Electron The Hall Effect Mass Spectrometer

27-1 Magnets and Magnetic Fields
Magnets have two ends – poles – called north and south. Like poles repel; unlike poles attract. Figure Like poles of a magnet repel; unlike poles attract. Red arrows indicate force direction.

However, if you cut a magnet in half, you don’t get a north pole and a south pole – you get two smaller magnets. Figure If you split a magnet, you won’t get isolated north and south poles; instead, two new magnets are produced, each with a north and a south pole.

Magnetic fields can be visualized using magnetic field lines, which are always closed loops. Figure (a) Visualizing magnetic field lines around a bar magnet, using iron filings and compass needles. The red end of the bar magnet is its north pole. The N pole of a nearby compass needle points away from the north pole of the magnet. (b) Magnetic field lines for a bar magnet.

The Earth’s magnetic field is similar to that of a bar magnet. Note that the Earth’s “North Pole” is really a south magnetic pole, as the north ends of magnets are attracted to it. Figure The Earth acts like a huge magnet; but its magnetic poles are not at the geographic poles, which are on the Earth’s rotation axis.

A uniform magnetic field is constant in magnitude and direction. The field between these two wide poles is nearly uniform. Figure Magnetic field between two wide poles of a magnet is nearly uniform, except near the edges.

27-2 Electric Currents Produce Magnetic Fields
Experiment shows that an electric current produces a magnetic field. The direction of the field is given by a right-hand rule. Figure (a) Deflection of compass needles near a current-carrying wire, showing the presence and direction of the magnetic field. (b) Magnetic field lines around an electric current in a straight wire. (c) Right-hand rule for remembering the direction of the magnetic field: when the thumb points in the direction of the conventional current, the fingers wrapped around the wire point in the direction of the magnetic field. See also the Chapter-Opening photo.

27-2 Electric Currents Produce Magnetic Fields
Here we see the field due to a current loop; the direction is again given by a right-hand rule. Figure Magnetic field lines due to a circular loop of wire. Figure Right-hand rule for determining the direction of the magnetic field relative to the current.

27-3 Force on an Electric Current in a Magnetic Field; Definition of B
A magnet exerts a force on a current-carrying wire. The direction of the force is given by a right-hand rule. Figure (a) Force on a current-carrying wire placed in a magnetic field B; (b) same, but current reversed; (c) right-hand rule for setup in (b).

The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation: This equation defines the magnetic field B. In vector notation:

Unit of B: the tesla, T: 1 T = 1 N/A·m. Another unit sometimes used: the gauss (G): 1 G = 10-4 T.

Example 27-1: Magnetic Force on a current-carrying wire. A wire carrying a 30-A current has a length l = 12 cm between the pole faces of a magnet at an angle θ = 60°, as shown. The magnetic field is approximately uniform at T. We ignore the field beyond the pole pieces. What is the magnitude of the force on the wire? Solution: F = IlBsin θ = 2.8 N.

Example 27-2: Measuring a magnetic field. A rectangular loop of wire hangs vertically as shown. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of wire ab (length l = 10.0 cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward magnetic force (in addition to the gravitational force) of F = 3.48 x 10-2 N when the wire carries a current I = A. What is the magnitude of the magnetic field B? Solution: The wire is perpendicular to the field (the vertical wires feel no force), so B = F/Il = 1.42 T.

Example 27-3: Magnetic Force on a semicircular wire. A rigid wire, carrying a current I, consists of a semicircle of radius R and two straight portions as shown. The wire lies in a plane perpendicular to a uniform magnetic field B0. Note choice of x and y axis. The straight portions each have length l within the field. Determine the net force on the wire due to the magnetic field B0. Solution: The forces on the straight sections are equal and opposite, and cancel. The force dF on a segment dl of the wire is IB0R dφ and is perpendicular to the plane of the diagram. Therefore, F = ∫dF = IB0R∫sin φ dφ = 2IB0R.

27-4 Force on an Electric Charge Moving in a Magnetic Field
The force on a moving charge is related to the force on a current: Once again, the direction is given by a right-hand rule. Figure Force on charged particles due to a magnetic field is perpendicular to the magnetic field direction.

Conceptual Example 27-4: Negative charge near a magnet. A negative charge -Q is placed at rest near a magnet. Will the charge begin to move? Will it feel a force? What if the charge were positive, +Q? Solution: There is no force on a motionless charge, be it positive or negative, so in neither case will it begin to move.

Example 27-5: Magnetic force on a proton. A magnetic field exerts a force of 8.0 x N toward the west on a proton moving vertically upward at a speed of 5.0 x 106 m/s (a). When moving horizontally in a northerly direction, the force on the proton is zero (b). Determine the magnitude and direction of the magnetic field in this region. (The charge on a proton is q = +e = 1.6 x C.) Solution: Since the force is zero when the proton is moving north, the field must point in the north-south direction. In order for the force to be to the west when the proton is moving up, the field must point north. B = F/qv = 0.10 T.

If a charged particle is moving perpendicular to a uniform magnetic field, its path will be a circle. Figure Force exerted by a uniform magnetic field on a moving charged particle (in this case, an electron) produces a circular path.

Example 27-7: Electron’s path in a uniform magnetic field. An electron travels at 2.0 x 107 m/s in a plane perpendicular to a uniform T magnetic field. Describe its path quantitatively. Solution: The magnetic force keeps the particle moving in a circle, so mv2/r = qvB. Solving for r gives r = mv/qB = 1.1 cm.

Conceptual Example 27-8: Stopping charged particles. Can a magnetic field be used to stop a single charged particle, as an electric field can? Solution: No, because the force is always perpendicular to the velocity. In fact, a uniform magnetic field cannot change the speed of a charged particle, only its direction.

Problem solving: Magnetic fields – things to remember: The magnetic force is perpendicular to the magnetic field direction. The right-hand rule is useful for determining directions. Equations in this chapter give magnitudes only. The right-hand rule gives the direction.

Conceptual Example 27-9: A helical path. What is the path of a charged particle in a uniform magnetic field if its velocity is not perpendicular to the magnetic field? Solution: The path is a helix – the component of velocity parallel to the magnetic field does not change, and the velocity in the plane perpendicular to the field describes a circle.

The aurora borealis (northern lights) is caused by charged particles from the solar wind spiraling along the Earth’s magnetic field, and colliding with air molecules. Figure (a) Diagram showing a negatively charged particle that approaches the Earth and is “captured” by the magnetic field of the Earth. Such particles follow the field lines toward the poles as shown. (b) Photo of aurora borealis (here, in Kansas, a rare sight).

Conceptual Example 27-10: Velocity selector, or filter: crossed E and B fields. Some electronic devices and experiments need a beam of charged particles all moving at nearly the same velocity. This can be achieved using both a uniform electric field and a uniform magnetic field, arranged so they are at right angles to each other. Particles of charge q pass through slit S1 and enter the region where B points into the page and E points down from the positive plate toward the negative plate. If the particles enter with different velocities, show how this device “selects” a particular velocity, and determine what this velocity is. Figure 27-21: A velocity selector: if v = E/B, the particles passing through S1 make it through S2. Solution: Only the particles whose velocities are such that the magnetic and electric forces exactly cancel will pass through both slits. We want qE = qvB, so v = E/B.

27-5 Torque on a Current Loop; Magnetic Dipole Moment
The forces on opposite sides of a current loop will be equal and opposite (if the field is uniform and the loop is symmetric), but there may be a torque. The magnitude of the torque is given by Figure Calculating the torque on a current loop in a magnetic field B. (a) Loop face parallel to B field lines; (b) top view; (c) loop makes an angle to B, reducing the torque since the lever arm is reduced.

The quantity NIA is called the magnetic dipole moment, μ: The potential energy of the loop depends on its orientation in the field:

Example 27-11: Torque on a coil. A circular coil of wire has a diameter of 20.0 cm and contains 10 loops. The current in each loop is 3.00 A, and the coil is placed in a 2.00-T external magnetic field. Determine the maximum and minimum torque exerted on the coil by the field. Solution: The minimum torque is zero; the maximum is NIAB sin  = 1.88 N·m.

Example 27-12: Magnetic moment of a hydrogen atom. Determine the magnetic dipole moment of the electron orbiting the proton of a hydrogen atom at a given instant, assuming (in the Bohr model) it is in its ground state with a circular orbit of radius r = x m. [This is a very rough picture of atomic structure, but nonetheless gives an accurate result.] Solution: The centripetal force is provided by the Coulomb force between the electron and proton: mv2/r = e2/4πε0r2. This can be solved to find the velocity of the electron: v = 2.19 x 106 m/s. The time required for one orbit is T = 2πr/v; the current is I = e/T. Finally, the dipole moment is μ = IA = 9.27 x J/T.

27-6 Applications: Motors, Loudspeakers, Galvanometers
An electric motor uses the torque on a current loop in a magnetic field to turn magnetic energy into kinetic energy. Figure Diagram of a simple dc motor. Figure The commutator-brush arrangement in a dc motor ensures alternation of the current in the armature to keep rotation continuous. The commutators are attached to the motor shaft and turn with it, whereas the brushes remain stationary.

Loudspeakers use the principle that a magnet exerts a force on a current-carrying wire to convert electrical signals into mechanical vibrations, producing sound. Figure Loudspeaker.

A galvanometer takes advantage of the torque on a current loop to measure current; the spring constant is calibrated so the scale reads in amperes. Figure Galvanometer.

27-7 Discovery and Properties of the Electron
Electrons were first observed in cathode ray tubes. These tubes had a very small amount of gas inside, and when a high voltage was applied to the cathode, some “cathode rays” appeared to travel from the cathode to the anode. Figure Discharge tube. In some models, one of the screens is the anode (positive plate).

The value of e/m for the cathode rays was measured in 1897 using the apparatus below; it was then that the rays began to be called electrons. Figure goes here. Figure Cathode rays deflected by electric and magnetic fields.

Millikan measured the electron charge directly shortly thereafter, using the oil-drop apparatus diagrammed below, and showed that the electron was a constituent of the atom (and not an atom itself, as its mass is far too small). The currently accepted values of the electron mass and charge are m = 9.1 x kg e = 1.6 x C Figure Millikan’s oil-drop experiment.

27-8 The Hall Effect When a current-carrying wire is placed in a magnetic field, there is a sideways force on the electrons in the wire. This tends to push them to one side and results in a potential difference from one side of the wire to the other; this is called the Hall effect. The emf differs in sign depending on the sign of the charge carriers; this is how it was first determined that the charge carriers in ordinary conductors are negatively charged. Figure The Hall effect. (a) Negative charges moving to the right as the current. (b) Positive charges moving to the left as the current.

27-8 The Hall Effect Example 27-13: Drift velocity using the Hall effect. A long copper strip 1.8 cm wide and 1.0 mm thick is placed in a 1.2-T magnetic field. When a steady current of 15 A passes through it, the Hall emf is measured to be 1.02 μV. Determine the drift velocity of the electrons and the density of free (conducting) electrons (number per unit volume) in the copper. Solution: The drift velocity is EH/Bd = 4.7 x 10-5 m/s. Since I = nevdA, n = I/evdA = 11 x 1028 m-3.

27-9 Mass Spectrometer A mass spectrometer measures the masses of atoms. If a charged particle is moving through perpendicular electric and magnetic fields, there is a particular speed at which it will not be deflected, which then allows the measurement of its mass:

27-9 Mass Spectrometer All the atoms reaching the second magnetic field will have the same speed; their radius of curvature will depend on their mass. Figure Bainbridge-type mass spectrometer. The magnetic fields B and B’ point out of the paper (indicated by the dots), for positive ions.

27-9 Mass Spectrometer Example 27-14: Mass spectrometry.
Carbon atoms of atomic mass 12.0 u are found to be mixed with another, unknown, element. In a mass spectrometer with fixed B′, the carbon traverses a path of radius 22.4 cm and the unknown’s path has a 26.2-cm radius. What is the unknown element? Assume the ions of both elements have the same charge. Solution: The ratio of masses is equal to the ratio of the radii, or Therefore the unknown element has a mass of 14.0 u, and is probably nitrogen.

Summary of Chapter 27 Magnets have north and south poles.
Like poles repel, unlike attract. Unit of magnetic field: tesla. Electric currents produce magnetic fields. A magnetic field exerts a force on an electric current:

Summary of Chapter 27 A magnetic field exerts a force on a moving charge: Torque on a current loop: Magnetic dipole moment:

Chapter 26 DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram.

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Chapter 26 DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram.

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Presentation on theme: "Chapter 26 DC Circuits Chapter 26 Opener. These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram."— Presentation transcript:

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