A 2. The potential difference across the capacitor is F m N m /C m. R R m m R m R m 0

Transcription

1 Model: The charged metal spheres are isolated and far from each other and anything else. Solve: (a) The charge on a sphere of radius r, charged to a potential V 0 is given by the equations: The ratio of charge densities is Q 4 0rV 0 Q 40rV 0 Q A Q A r 4r r R Q Q A r 4r r R A (b) The electric field strengths at the surface of the conductors is E E and E, thus. 0 0 E Assess: Halving the radius doubles the electric field strength as well as the charge density. Sharp corners (i.e., parts of conductors with a small radius of curvature, like a small sphere) have strong electric fields. This characteristic is why lightning rods work Model: Capacitance is a geometric property of two electrodes. Solve: The ratio of the charge to the potential difference is called the capacitance: C Q V. C The potential difference across the capacitor is Using R R.0 mm, Q Q Q VC 4 0 R 4 0 R 4 0 R R RR RR C R R R R.00 m 0 F RR 000 F.00 m 9.00 N m /C 9000 m R R.00 m 9000 m R.00 m R 9000 m m.00 m m R m.95 cm The outer radius is R R 0.00 m m 3.05 cm. So, the diameters are 5.9 cm and 6. cm Model: Assume the battery is ideal.

2 The circuit in the figure has been redrawn to show that the six capacitors are arranged in three parallel combinations, each combination being a series combination of two capacitors. Solve: (a) The equivalent capacitance of the two capacitors in series is C. The equivalent capacitance of the six capacitors is 3 C. (b) As points a and b are midpoints of identical capacitors, V a V b 6.0 V. Therefore, the potential difference between points a and b is zero Model: Assume the battery is ideal. Solve: When the capacitors are individually charged, their charges are Q CV 0 F 0 V 00 C Q CV 0 F 0 V 00 C These two capacitors are then connected with the positive plate of C connected with the negative plate of C as shown in the figure. Let the new charges on C and C be Q and Q. Then, Q Q 00 C 00 C 00 C The voltages across parallel capacitors are the same, so Q Q C V V Q Q C C C Substituting this expression for Q into the previous equation, C 0 F Q Q 00 C Q 00 C C 0 F Solving these equations, we get Q 33 C and Q 67 C. Finally, Q 33 C V 3.3 V V C 0 F Solve: (a) From Table 3., the resistivity of aluminum is m. From the equation for the resistivity, the length L of a wire with a cross-sectional area A and having a resistance R is 6 00 m 000 AR L 3.57 m 3.6 m 8.80 m (b) The number of turns is the length of the wire divided by the circumference of one turn. Thus, 3.57 m m Solve: (a) A current of.8 pa for the potassium ions means that a charge of.8 pc flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is.80 C/s 9.60 C.50 s 7

3 Since the channel opens only for.0 ms, the total number of potassium ions that pass through the channel is.5 0 s.0 0 s atoms. (b) The current density in the ion channel is.8 pa.80 A I J.50 A/m A nm 0.50 m Solve: The current density is J I A. This means Assess: Fuse wires are usually thin A A D I D I cm 0.50 mm 4 J J 500 A/cm Solve: (a) The charge delivered is A 500 s.5 C. (b) The current in the lightning rod and the potential drop across it are related by Equation 3.. Using for iron from Table 3., m5.0 m500 3 A A LI 4 I V A.430 m L V 00 V This is the area required for a maximum voltage drop of 00 V. The corresponding diameter of the lightning rod is A r m m 8.8 mm Please refer to the figure in the text. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulb s resistance R is in series with the internal resistance r, giving a total resistance R eq R r. The current is.50 V I E bat 0.3 A Rr 6.50 This is the current leaving the battery. But all of this current flows through bulb A, so I A I bat 0.3 A. (b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance R eq R Their equivalent resistance is in series with the battery s internal resistance, so the current flowing from the battery is.50 V I E bat 0.48 A R r 3.50 eq But only half this current goes through bulb A, with the other half through bulb B, so IA I bat 0.4 A. (c) The change in I A when the switch is closed is 0.07 A. This is a decrease of 7.4%. (d) If r 0, the current when the switch is open would be I A I bat 0.50 A. With the switch closed, the current would be I bat A and the current through bulb A would be IA I bat 0.50 A. The current through A would not change when the switch is closed.

4 3.63. Model: The batteries and the connecting wires are ideal. The figure shows how to simplify the circuit using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and build up the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the figure and from Kirchhoff s loop law, V 3 V I.0 A 9 Thus, the current through the batteries is.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. In Step of the above figure, the 9 resistor is returned to the 6 and 3 resistors in series. Both resistors must have the same.0 A current as the 6 resistor. We use Ohm s law to find V 3 (.0 A)(3 ) 3.0 V V V As a check, 3.0 V 6.0 V 9 V, which was V ( V 3 V) 9 V of the 9 resistor. In Step, the 6 resistor is returned to the 4 and 8 resistors in parallel. The two resistors must have the same potential difference V 6.0 V. From Ohm s law, 6.0 V 3 I7 A V I4 A 4 4

5 As a check, 0.75 A 0.5 A.0 A which was the current I of the 6 resistor. In Step 3, the 8 resistor is returned to the 3 and 5 (right) resistors in series, so the two resistors must have the same current of 0.88 A. We use Ohm s law to find V 3 (3/4 A)(3 ) 9/4 V V 4 (3/4 A)(5 ) 5/4 V As a check, 9/4 V 5/4 V 4/4 V = 6.0 V, which was V of the 8 resistor. In Step 4, the 3 resistor is returned to 4 (left) and resistors in parallel, so the two must have the same potential difference V 9/4 V. From Ohm s law, 9/4 V I4 9/6 A = 0.56 A 4 9/4 V 9 I A 0.9 A 48 As a check, 0.56 A 0.9 A 0.75 A, which was the same as the current through the 3 resistor. Resistor Potential difference (V) Current (A) Model: The wires are ideal, but the batteries are not. Solve: (a) The good battery alone can drive a current through the starter motor V I 00 A (b) Alone, the dead battery drives a current 8.0 V I A (c) Let I, I, I 3 be defined as shown in the figure above. Kirchhoff s laws applied to the good battery and dead battery loop, good battery and starter motor loop, and the top middle junction yield three equations in the three unknown currents: V I0.0 I V I0.0 I V 0 I I I 3 Substituting for I from the third equation into the first and second equations gives V I0.0 I V I0.5 I Solving for I from the first equation, V I I 0.0 Substituting into the second equation and solving for I 3 yields the current through the starter motor is 99 A. (d) Substituting the value for I 3 into the expression for I yields the current through the dead battery is 3.9 A. Assess: The good battery is charging the dead battery as well as running the started motor. A total of 03 A flows through the good battery.

6 3.73. Model: The battery and the connecting wires are ideal. Please refer to the figure in the text Solve: (a) A very long time after the switch has closed, the potential difference V C across the capacitor is E. This is because the capacitor charges until V C E, while the charging current approaches zero. (b) The full charge of the capacitor is Q max C(V C ) max C E. (c) In this circuit, I dq dt because the capacitor is charging, that is, because the charge on the capacitor is increasing. (d) From Equation 3.36, capacitor charge at time t is Q Q max ( e t/ ). So, A graph of I as a function of t is shown below. dq d t t t E I CE e CE e CE e e dt dt RC R t Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Please refer to Figure EX The current in the circuit on the left is I and has a clockwise direction. The current in the circuit on the right is I and has a counterclockwise direction. Solve: Since I 9 V 4.5 A, the force between the two wires is F N 0 7 T m/a0.0 m4.5 A LI I d m 9 V I 3.0 A R A I Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Please refer to the figure in the text Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v. Upon entering the magnetic field, the ion follows a circular trajectory with cyclotron radius r mv/eb. To be detected, an ion s trajectory must have radius d r 8 cm. This means the ion needs the speed ebr ebd v m m

7 This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy equation to find the voltage that will accelerate the ion: mv K U K U 0 J ev mv 0 J V e Using the above expression for v, the voltage that causes an ion to be detected is mv m ebd eb d V e e m 8m An ion s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of is N m m N m N m elec (4.003 u)( kg/u) kg kg Note that we re given the atomic masses very accurately in Exercise 8. We need to retain this accuracy to tell the difference between and CO. The voltage for N is N C0.00 T 0.08 m kg V 0.07 V Ion Mass (kg) Accelerating voltage (V) N O CO Assess: The difference between N and CO is not large but is easily detectable Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop balances the torque of the weight. Please refer to Figure P Solve: The rotational equilibrium condition net 0 N m is about the axle and means that the torque from the weight is equal and opposite to the torque from the magnetic force. We have kg g 0.05 m Bsin90 NIA kg 9.8 m/s 0.05 m B 0.3 T 0.0 A0.050 m0.00 m Assess: The current in the loop must be clockwise for the two torques to be equal. B Model: The bar is a current-carrying wire in a perpendicular uniform magnetic field. The current is constant. Please refer to the figure in the text. Solve: (a) The right-hand rule as described in section 33.8 requires the current to be into the page. (b) The net force on the bar is F IlB, and is constant throughout the motion. The acceleration of the bar is F IlB thus. Using constant acceleration kinematics, m m IlB IlBd vf v0 as 0 m/s d vf m m Model: Assume the wire is infinitely long. See Figure P The current through the wire produces a steady magnetic field, but the magnetic field strength depends on the distance from the wire. The loop moves away from the wire at constant speed. The

8 flux through the loop varies due to its motion. Faraday s law gives the magnitude of the induced emf, and Ohm s law will yield the current strength. The direction of the magnetic field is into the paper at the location of the loop. The magnetic field is perpendicular to the plane of the loop, so BdA BdA. Solve: The flux through the loop is decreasing as it moves away from the wire. Lenz s law implies that the induced current is clockwise in order to increase the flux. Using the results of Example 34.5, which treats the rectangular loop as a series of tall infinitesimally thin strips, 0I(4.0 cm) x.0 cm m ln x where x is the distance from the wire to the closer edge of the loop, and changes with time. Note dm dm dx dm dx v, where v is the velocity of the loop. dt dx dt dx dt The induced emf is thus At x =.0 cm, dm dx 0 I(4.0 cm).0 cm E v dx dt x x.0 cm I.0 cm 0 6 E 4.0 cm 0 m/s.3 0 V.0 cm 3.0 cm Since the loop resistance is R = 0.00, the induced current is thus E Iloop 6.70 R Assess: The induced emf is proportional to the current in the wire. For reasonable currents (~ A) the induced emf is reasonable. 5 I Model: Assume the magnetic field is uniform over the plane of the loop. The oscillating magnetic field strength produces a changing flux through the loop and an induced emf in the loop. Solve: (a) The normal to the surface of the loop is in the same direction as the magnetic field so that ABBA. The induced emf is d db db E A r = r B0 cost dt dt dt The cosine will oscillate between and so the maximum emf is E r B r f B Hz 00 T 0.93 V 6 9 max 0 0 (b) If the loop is rotated so that the plane is perpendicular to the electric field, then the normal to the surface will be parallel to the magnetic field. There is no magnetic flux through the loop and no induced emf Model: Assume negligible resistance in the LC part of the circuit. With the switch in position for a long time the capacitor is fully charged. After moving the switch to position, there will be oscillations in the LC part of the circuit. Solve: (a) After a long time the potential across the capacitor will be that of the battery and Q0 CVb att. When the switch is moved, the capacitor will discharge through the inductor and LC oscillations will begin. The maximum current is H CVbatt C.0 0 F 0 0 batt batt 3 V A 76 ma I Q C V V LC L (b) The current will be a maximum one-quarter cycle after the maximum charge. The period is

9 3 6 T LC 500 H.00 F.0 ms So the current is first maximum at tmax T 0.50 ms Model: Assume an ideal inductor and an ideal (resistanceless) battery. Please refer to the figure in the text. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference V L L(dI/dt) would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Consequently, the inductor has no effect on the circuit. It is simply a 0 resistor and 0 resistor in series with the battery. The equivalent resistance is 30, so the current through the circuit (including through the 0 resistor) is I V bat /R eq (30 V)/(30 ).0 A. (b) After a long time, the currents in the circuit will reach steady values and no longer change. With steady currents, the potential difference across the inductor is V L L(dI/dt) 0 V. An ideal inductor has no resistance (R 0 ), so the inductor simply acts like a wire. In this case, the inductor shorts out the 0 resistor. All current from the 0 resistor flows through the resistanceless inductor, so the current through the 0 resistor is 0 A. (c) When the switch has been closed a long time, and the inductor is shorting out the 0 resistor, the current passing through the 0 resistor and through the inductor is I (30 V)/(0 ) 3.0 A. Because the current through an inductor cannot change instantaneously, the current must remain 3.0 A immediately after the switch reopens. This current must go somewhere (conservation of current), but now the open switch prevents the current from going back to the battery. Instead, it must flow upward through the 0 resistor. That is, the current flows around the LR circuit consisting of the 0 resistor and the inductor. This current will decay with time, with time constant L/R, but immediately after the switch reopens the current is 3.0 A Model: Use the Galilean transformation of fields. A current of.5 A flows to the right through the wire, and the plastic insulation has a charge of linear density.5 n C/cm. Solve: The magnetic field B at a distance r from the wire is 0 I B, clockwise seen from left r On the other hand, E is radially out along ˆr, that is, E rˆ r As the mosquito is.0 cm from the center of the wire at the top of the wire, 0

10 7 4 0 T m/a.5 A 0.00 m B T.5 0 C/m N m /C 5 V E m m where the direction of B is out of the page and the direction of E is radially outward. In the mosquito s frame (let us call it S), we want B 0 T. Thus, B B V E 0 B V E c c Because V E must be in the direction of B and E is radially outward, according to the right-hand rule V must be along the direction of the current. The magnitude of the velocity is cb m/s T V 5 E 4.50 V/m m/s The mosquito must fly at m/s parallel to the current. This is highly unlikely to happen unless the mosquito is from Planet Krypton, like Superman Model: The microwave beam is an electromagnetic wave. The water does not lose heat during the process. Solve: The rate of energy transfer from the beam to the cube is C 0 P0.80IA0.80 E0 A 8 30 m/s C /Nm V/m 0.0 m.9 kw The amount of energy required to raise the temperature by 50 C is E mct 0.0 m 000 kg/m 486 J/kg/ C50 C.09 0 J The time required for the water to absorb this much energy from the microwave beam is 5 E.090 J t 6 s 3 P.90 W Assess: Raising kg of water by 50 C in a microwave oven takes around 3 minutes, so this is reasonable Model: Use the particle model for the astronaut. Solve: According to Newton s third law, the force of the radiation on the astronaut is equal to the momentum delivered by the radiation. For this force we have P 000 W F prad A c m/s Using Newton s second law, the acceleration of the astronaut is N

11 N m/s a 80 kg Using v f v i a(t f t i ) and a time equal to the lifetime of the batteries, v f 0 m/s ( m/s )(3600 s) m/s The distance traveled in the first hour is calculated as follows: v v a s f i first hour ( m/s) (0 m/s) ( m/s )(s) first hour (s) first hour 0.70 m This means the astronaut must cover a distance of 5.0 m 0.7 m 4.73 m in a time of 9 hours. The acceleration is zero during this time. The time it will take the astronaut to reach the space capsule is 4.73 m t 3,533 s 8.76 hours m/s Because this time is less than 9 hours, the astronaut is able to make it safely to the space capsule Model: Use Malus s law for the polarized light. Solve: For unpolarized light, the electric field vector varies randomly through all possible values of. Because the average value of cos is, the intensity transmitted by a polarizing filter when the incident light is unpolarized is I I 0. For polarized light, I transmitted I 0 cos. Therefore, I I cos 45 I 3 I cos 45 I 3 (I cos 45 )cos 45 I 0(cos 4 45 ) I 8 0