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course PHY 202
4/24 9 pmSome of these are questions on my last set of questions and the rest are new.
Problem Number 2
Under certain conditions of temperature and potential difference, a uniform wire with cross-sectional radius 3.6 cm carries a current of 1.7 amps. Under the same conditions, what will be the current of a second wire whose length is identical to that of the first, but which has diameter 6.3 cm?
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.V/.0036 m and V/.0063 m so the potential gradient is V/.0063 m /V/.0036 m= .0036m/.0063 m=.57 times greater than the first wire
Therefore the current is .57 *1.7 amps=.969 amps
@& Potential gradient is change in potential / distance. In this case the potential gradient would be the voltage divided by 3.6 cm. Both wires have the same length and therefore the same potential gradient.
The potential gradient is the force per unit of charge. It determines how much force is accelerating the electrons. This acceleration is the same in both wires.
The difference is in the cross-sectional areas. The more cross-sectional area, the more electrons are available to be accelerated, and the greater the current.
The question is therefore how the cross-sectional areas of the two wires compare. *@
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Was I semi-close in my answer? I did compare them, but was it in the right way.
@& The potential gradient is potential difference / length.
Potential difference and length are the same, so potential gradient is the same.
You didn't compare the cross-sectional areas. That is the essential comparison.
The ratio of the diameters is (6.3 cm / (3.6 cm) ).
The ratio of the areas is therefore (6.3 cm / (3.6 cm) )^2, approximately 3.
Since the second wire has three times the cross-sectional area, it has three times as many electrons available, per unit of length, to be accelerated. So it will carry about 3 times the current.
If aren't sure about the ratios, you can calculate the cross-sectional areas, knowing the two diameters.*@
Problem set 2/2
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Problem Number 6
A sensitive balance shows that a force of .0087 Newtons is exerted on a straight wire 3 cm long carrying a current of 5.7 amps by a uniform magnetic field directed perpendicular to the wire. What is the strength of the field? If the force is to the East and the current runs to the South, what is the direction of the field?
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F=.0087 N x=.03 m current 5.7 amps
B=9 * 10^-7 T m / amp(5.7 amps*.03m/.015 m^2)
I know this is not completely right. How do I find the field strength without an angle, or is the angle 90 degrees?
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@& The field is perpendicular to the wire, so the angle is 90 degrees.
The right-hand rule determines the direction of the field. The field and the current are perpendicular, and the force is perpendicular to both. So the field is perpendicular to the force (and, as is given, to the current).
What directions are therefore candidates for the field?
Which of these directions is confirmed by the right-hand rule?*@
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Would the direction of the field be upward by the right hand rule? If my fingers point east, then south, then my thumb is pointed upward.
@& Fingers held straight out point in the direction of the current, fingers when curled 90 degrees relative to the palm represent the direction of the magnetic field, thumb in the direction of the force.
So thumb is to the east, fingers south, and when you curl your fingers then point downward.
The field is therefore downward.*@
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Problem Number 2
A coil consists of 56 circular loops of wire each of radius .14 meters. A uniform magnetic field with strength .0024 Tesla is rotated in such a way that at t = 0 the field is perpendicular to the loop, while at t = .008 sec the field makes an angle of 63 degrees with a perpendicular to the plane of the loop. What is the average voltage induced around the coil?
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area=(56) * `pi (.14 m)^2=3.45 m^2
3.45 m^2 * .0024 Tesla cos (63)= 3.76 Tesla*m^2
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@& You've got the area. Multiply it by the whole field to get the flux when perpendicular. Your calculation gives the flux when the angle is 63 degrees.
So what is the change in flux, and what is the change in clock time?
What therefore is the average rate of change of flux with respect to clock time?
That's the average voltage.*@
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Problem Number 3
What force is exerted on a straight wire 6 cm long, carrying current 4.3 amps, by a uniform magnetic field of .0037 Tesla oriented perpendicular to the wire? If the current runs North and the magnetic field runs to the East, what is the direction of the force?
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F= .06 m * 4.3 amps * .0037 Tesla * sin90=9.54 x10^-4 N and the direction of the force would be downward.
@& right*@
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Problem Number 1
Answer the following:
If a potential gradient of 16 volts/meter exists between 2 point separated by .75 meters parallel to the gradient, what is the voltage between the 2 points?
How much work will be done on a charge of 91 Coulombs as it moves between these points?
What therefore is the average force on the charge?
Repeat these calculations for the same potential gradient with a displacement of 3.6 meters.
From your results, state whether it appears to be the potential gradient or the charge which determines the force.
From potential gradient dV/dx and displacement `ds, find an expression for the work done on a charge q as it moves through this displacement, and use this expression to find an expression for the force on the charge.
Give the resulting force for a charge of 43 Coulombs within a potential gradient of `potGrad 2 volts/meter.
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Voltage=16 volts/meter *.75 m=12 V
Electric field= 16 V/N * 91 C=1456 N
W=1456N* .75 m=1092 J
Fave=1092 J/.75 m=1456 N
Voltage=16 meter/volt*3.6 m=57.6 V
W=1456 N *3.6 m=5241.6 J
Fave=5241.6 J/3.6=1456 N
According to my results, the potential gradient determines the force.
W=dV/dx * `ds * Q
F*ds=dW* Q
Fave=43 C * 2 volts/meter=86 N
@##$%
@& Good.*@
Problem Number 2
A generator has a negligible internal resistance. It creates a potential difference of 4.7 volts when cranked at 1 rev/sec, and the voltage is proportional to the cranking rate. Ther generator cranked at 3.36 rev/s and is connected to a series combination of two bulbs, one with a resistance of 4.7 ohms and the other with resistance 23 ohms. How much current flows in each bulb? How much power is dissipated in each?
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Voltage=3.6 rev/sec * 4.7 V= 16.92 V
I1=16.92 V/ 4.7 ohms=3.6 C
P1=3.6 C*16.92 V=60.9 Watts
I2=16.92 V/23 ohms=.74 C
P2=.74 C*16.92 V= 12.45 Watts
In a previous question I answered, you said that the voltage drops equal the sum of the voltage. How do you determine the voltage in each if you’re given solely the resistance?
@& Assume current I.
The voltage drop in the first resistor is I * 4.7 ohms.
The drop in the second is I * 23 ohms.
The total drop is therefore
I * 4.7 ohms + I * 23 ohms = 16.9 volts.
Solve for I.*@
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Problem Number 6
How much power is required to maintain the current that results when a potential difference of 4 Volts is maintained across a resistance of 26 ohms?
I=4 Volts/ 26 ohms=.15 C
Power=.15 C*4 V=.6 Watts
@& Good.*@
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Problem Number 7
Find the magnetic flux through a square loop with side .28 meters, in the presence of a magnetic field with strength .0081 Tesla, when the field is perpendicular to the loop.
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Area is .28 m ^2=.078 m^2
F= .00081 Tesla * .078 m^2= .00063 Tesla m^2
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Problem Number 1
Find the magnitude of the force between charges of .0006 Coulombs at (-7.008 m, -4.005 m) and .0004 Coulombs at (-9.01 m, 8 m) on the x-y plane.
College and University Physics Students: Find the direction of the force exerted on the second charge by the first, and the vector describing this force.
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Distance between charges= r=sqrt (-9.07 m- (-7.008 m)) + (8 M - (-4.005 m))=
R=sqrt (-2.06 m + 12.005 m)=sqrt (9.945 m)
r=3.15
F= 9 * 10^9 N m^2 / C^2* .0006 C * .0004 C/ (3.15 m)^2 =2160
@& Good so far.
A triangle from the first point to the second has legs making displacements of -2 meters and +12 meters, respectively, in the x and y directions.
The corresponding angle is
arcTan ( 12 m / (-2 m) ) = arcTan(-6),
about -84 degrees with the x axis. This puts the direction at about 180 deg - 84 deg = 96 deg.
The charges are like so the force is repulsive, and the 2160 N force acts at 96 deg.
Its x and y components are
2160 N * cos(96 deg) = -200 N, approx.
and
2160 N * sin(96 deg) = 2100 N, approx.*@
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Problem Number 5
An electron (mass 9.11 * 10^-31 kg) with velocity directed to the North passes through a magnetic field of .006 Tesla directed vertically upward, crossed with an electric field of 42000 N/C directed either East or West. The electron passes through undeflected. Is the electric field directed East or West, and how fast is the electron moving? What force does the electron experience from each field?
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r = v * m / q
Without given the charge, how can we solve for v?
Then using v, the force= B V q
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@& The electrostatic force on the electron is q E.
The magnetic force is q v B.
These forces are equal and opposite if the path of the electron is not changed.
So the magnitudes of the two forces are equal:
q * E = q * v * B.
Solve for v.
The charge q divides out, and v = E / B.
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@& Check my notes. Glad to answer more quesitons.*@