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course Phy 242
April 12 around 1:30pm. I didn't get a lot of these answered. Had some trouble.
110323`q001. How much energy does it take to move a 2 microCoulomb charge from x = 30 cm to x = 10 cm, given a field of 8 * 10^5 Newtons / Coulomb directed in the positive x direction?
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F = Qe
= (8*10^5) * (2*10^-6)
= 1.6 N
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As shown in Class Notes the force exerted on the charge by the field is
F = 2 microCoulombs * (8 * 10^5 N / C) = 1.6 Newtons.
This commonsense idea is formulated as F = q * E.
To move the charge at a constant velocity toward the origin takes a force equal and opposite to the force exerted by the field, so the necessary force is in the direction of motion and the work will be positive.
The work is just `dW = ave force * displacement, so
`dW = -1.6 N * (10 cm - 30 cm) = 32 N * cm = 32 N * (.01 m) = .32 Joules.
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`q002. How much force is experienced by a 1 microCoulomb charge at the point (30 cm, 40 cm) by a 3 microCoulomb charge at the origin?
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r = <30, 40>, r = sqrt (30^2 + 40^2) = 50 cm
F = (9*10^9) * [(1 muC * 3 muC) / (0.5m^2)]
= 0.108 N
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What is the direction of this force?
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Northeast OR 0.108 N away from the origin
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How much work would it take to move the 1 microCoulomb charge one centimeter closer to the origin?
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(0.108 N) / (1*10^-6)
= 1.08*10^5 N/c (electric field)
WORK:
0.108 N * 1cm = 0.00108 J
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How many Joules does it take per Coulomb of moving charge to move that one centimeter?
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(0.00108 J) / (1*10^-6 C)
= 1080 J/C
= 1080 Volts
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What is the electric field at the point (30 cm, 40 cm) due to a 3 microCoulomb charge at the origin?
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`q003. A volt is a Joule of potential energy per Coulomb.
How much does chemical potential energy stored in a 1.5 volt battery change, per minute, as it drives a .20 amp current?
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0.2 amp = 0.2 C/s and 1.5 Volts = 1.5 J/C
0.2 * 1.5 J = 0.3 J
Requires = 0.30 J/s * 60 sec
= 18 J
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By how much does the electrostatic potential energy of a capacitor change in 200 milliseconds, if the capacitor is at 6 volts and drives a current of 50 milliamps?
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0.3 J/s * 0.2 sec
= 0.06 J
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How much energy does it take to light your lightbulb at 1.5 volts for 10 seconds?
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`q004. A magnet spins at 10 revolutions per second in the vicinity of a coil of wire enclosing a total area of 1000 cm^2. The magnet creates a maximum field of .04 Tesla within that area. At what average rate is the magnetic flux through the coil changing?
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The maximum magnetic flux is
phi_B = 1000 cm^2 * .04 T = .004 T * m^2.
The minimum flux is -.004 T * m^2.
The magnitude of flux change between max and min, or between min and max, is therefore .008 T * m^2.
This magnitude of flux change occurs twice per revolution, therefore 20 times per second. The corresponding time interval is .05 second.
So the average rate of flux change is approximately
.008 T * m^2 / (.05 second) = .160 T m^2 / sec = .16 volt.
The true average rate is found by integrating the instantaneous rate with respect to clock time, and differs slightly from this value.
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`q005. Charge Q creates electrostatic flux 4 pi k Q, with k = 9 * 10^9 N m^2 / Coulomb^2. A 5 meter length of wire contains a uniformly distributed charge of 8 microCoulombs.
How much charge does a 25 cm length of the wire contain?
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What will therefore be the electrostatic flux through a piece of pipe 25 cm long, which surrounds the wire?
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If the diameter of the pipe is 5 cm, then how much flux is there per square centimeter of the pipe?
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How much is that per square meter?
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5 meters of the wire contains 8 microCoulombs, so 25 cm of the wire contains .4 microCoulombs. This can be reasoned as follows:
.25 cm is 1/20 of 5 meters, so the charge enclosed is 1/20 of 8 microCoulombs, or .4 microCoulombs.
Alternatively, 8 microCoulombs on 5 meters of wires implies charge density 8 microCoulombs / (5 meters) = 1.6 microCoulombs / meter. So a .25 meter section has charge 1.6 microCoulombs / meter * .25 meter = .4 microCoulombs = 4 * 10^-7 C.
The electrostatic flux through the pipe is therefore 4 pi k Q = 4 pi * 9 * 10^9 N m^2 / C^2 * 4 * 10^-7 C = 14 400 pi N m^2 / C.
The flux exits radially through the circular surface of the pipe, so the area through which it exits is (circumference * length) = (2 pi * 5 cm) * (25 cm) = 250 pi cm^2 = .0025 pi m^2.
The flux / area is therefore
phi_E / area = (14 400 pi N m^2 / C) / (.0025 pi m^2) = 5 800 000 N / C,
approximately.
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`q006. Suppose the blackboard is uniformly covered by electrostatic charge, with a total charge of 60 microCoulombs. We place a piece of tile 1 cm from the board, parallel to and near the center of the board. The tile dimensions are 15 cm by 15 cm. The tile does not affect the electric field, which effectively passes right through it.
What must be the direction of the electric field relative to the surface of the tile?
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Half the flux is through the board, the other half is through the tile. By symmetry, the electric field is perp to the board. Straight out of the blackboard.
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What do you think is the electrostatic flux through the tile?
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Flux = field * area
(60 muC) / (20 tiles) approx
= 0.3 muC
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What is the electrostatic flux per square centimeter, and per square meter of the tile surface?
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4 pi kQ = (3*10^4 Nm^2/C)
Flux = 1.5*10^4 Nm^2/C
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How much do you think this result will change if we move the tile 2 cm further from the board?
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Field = flux/area
(1.5*10^4 Nm^2/C) / (15cm)^2
= 7*10^6 N/C
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`q007. A current of .25 amps is driven by a 1.5 volt battery through a 100-turn coil of radius 2 cm.
What magnetic field results at the center of the coil?
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What therefore is the magnetic flux within the coil?
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If the battery is removed the current in the coil will quickly fall to zero. If this occurs in 20 milliseconds, then at what average rate must the magnetic flux change?
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The magnetic field of a small segment wire of length `dL will be `dB = k ' * I `dL / r^2. By the right-hand rule the field will be perpendicular to the plane of the loop.
If we assume that the coil is pretty much circular (as opposed to cylindrical) the contributions of all segments reinforce, and the field is
B = k ' * I L / r^2,
where L is the total length of the wire.
Every 2 cm loop of wire has circumference 4 pi cm. There are 100 loops, so their total length is 400 pi cm.
Thus
B = 10^-7 T * m / amp * .25 amps * 400 pi cm / (2 cm)^2
= 10^-7 T * m / amp * .25 amps * 4 pi meter / (.02 m)^2
= 8 * 10^-4 T,
very approximately.
Note that the magnetic field of the Earth is on the order of 10^-4 Tesla. So this field would pretty seriously mess with a compass.
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Self-critique (if necessary):
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Self-critique rating:
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04-12-2011
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A current of .20 amps is .20 Coulombs / second. If the current lasts a minute, then the charge passing through the current in that time is .20 C / s * 60 s = 12 C.
1.5 volts means 1.5 Joules / Coulomb, so the loss of potential energy is
1.5 J / C * 12 C = 18 J.
A capacitor driving a current of 50 milliamps = .05 amps .05 C / s, for 200 milliseconds = .2 seconds, has charge .05 C / s * .2 s = .01 C passing through it in that time interval.
.01 C passing through a 6 volt increase in voltage would increase its PE by .01 C * 6 J / C = .06 J.
If for example the lightbulb you observed in class had a current of 100 milliamps at voltage 3 volts, then in 10 seconds the charge flowing through it would be 100 milliAmps * 3 volts = .1 C / s * 3 J / C = .3 Joules.
In symbols, these calculations are summarized by
`dQ = I * `dt and
| `dPE | = | `dQ * V |.
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5 meters of the wire contains 8 microCoulombs, so 25 cm of the wire contains .4 microCoulombs. This can be reasoned as follows:
.25 cm is 1/20 of 5 meters, so the charge enclosed is 1/20 of 8 microCoulombs, or .4 microCoulombs.
Alternatively, 8 microCoulombs on 5 meters of wires implies charge density 8 microCoulombs / (5 meters) = 1.6 microCoulombs / meter. So a .25 meter section has charge 1.6 microCoulombs / meter * .25 meter = .4 microCoulombs = 4 * 10^-7 C.
The electrostatic flux through the pipe is therefore 4 pi k Q = 4 pi * 9 * 10^9 N m^2 / C^2 * 4 * 10^-7 C = 14 400 pi N m^2 / C.
The flux exits radially through the circular surface of the pipe, so the area through which it exits is (circumference * length) = (2 pi * 5 cm) * (25 cm) = 250 pi cm^2 = .0025 pi m^2.
The flux / area is therefore
phi_E / area = (14 400 pi N m^2 / C) / (.0025 pi m^2) = 5 800 000 N / C,
approximately.
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@& Check the notes I've inserted.*@