If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
029. `Query 29
Question:
`qQuery introductory
problem set 54 #'s 8-13
Explain how to determine the magnetic flux of a uniform
magnetic field through a plane loop of wire, and explain how the direction of
the field and the direction of a line perpendicular to the plane of the region
affect the result.
Your Solution:
Confidence rating:
Given Solution:
To do this we need to simply find the area of the plane
loop of wire. If we are given the radius
we can find the area using
Pi * r
^2
Then we multiply the area of the loop (In square meters ) by
the strength of the field (in tesla).
This will give us the strength of the flux if the plane of
the loop is perpendicular to the field.
If the perpendicular to the loop is at some nonzero angle with the
field, then we multiply the previous result by the cosine of the angle.
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Question:
`qExplain how to determine the average rate of change of
magnetic flux due to a uniform magnetic field through a plane loop of wire, as
the loop is rotated in a given time interval from an orientation perpendicular
to the magnetic field to an orientation parallel to the magnetic field.
Your Solution:
Confidence rating:
Given Solution:
** EXPLANATION BY STUDENT:
The first thing that we need to do is again use Pi * r ^
2 to find the area of the loop. Then we multiply the area of the loop (m^2)
by the strength of the field (testla) to find the flux when the loop is
perpendicular to the field.
Then we do the same thing for when the loop is parallel to
the field, and since the cos of zero degrees is zero, the flux when the loop is
parallel to the field is zero. This
makes sense because at this orientation the loop will pick up none of the
magnetic field.
So now we have Flux 1 and Flux 2 being when the loop is
perpendicular and parallel, respectively.
So if we subtract Flux 2 from flux 1 and divide this value by the given
time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in
Tesla m^2 / sec = volts. **
Self-critique (if necessary):
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Question:
`qExplain how alternating current is produced by rotating a
coil of wire with respect to a uniform magnetic field.
Your Solution:
Confidence rating:
Given Solution:
** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y
ou rotate a coil of wire end over end inside a uniform
magnetic field. When the coil is
parallel to the magnetic field, then there is no magnetic flux, and the current
will be zero. But then when the coil is
perpendicular to the field or at 90 degrees to the field then the flux will be
strongest and the current will be moving in one direction. Then when the coil is parallel again at 180
degrees then the flux and the current will be zero. Then when the coil is perpendicular again at
270 degrees, then the flux will be at its strongest again but it will be in the
opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will
be moving in one direction and at 270 degrees the current will be moving with
the same magnitude but in the opposite direction.
COMMENT:
Good. The changing magnetic flux produces voltage,
which in turn produces current. **
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Question:
`qQuery Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance?
Your Solution:
Confidence rating:
Given Solution:
current = voltage / resistance (Ohm's Law). The common sense of this is that for a given
voltage, less resistance implies greater current while for given resistance,
greater voltage implies greater current.
More specifically, current is directly proportional to voltage and
inversely proportional to resistance. In
symbols this relationship is expressed as I = V / R.
In this case we know the current and the voltage and wish to
find the resistance. Simple algebra
gives us R = V / I. Substituting our
known current and voltage we obtain
R = 120 volts / 4.2 amps = 29 ohms, approximately.
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Question:
`qQuery Principles and General College Physics 18.28. Max instantaneous voltage to a 2.7 kOhm
resistor rated at 1/4 watt.
Your Solution:
Confidence rating:
Given Solution:
Voltage is energy per unit of charge, measured in Joules /
Coulomb.
Current is charge / unit of time, measured in amps or
Coulombs / second.
Power is energy / unit of time measured in Joules / second.
The three are related in a way that is obvious from the
meanings of the terms. If we multiply
Joules / Coulomb by Coulombs / second we get Joules / second, so voltage *
current = power. In symbols this is
power = V * I.
Ohm's Law tells us that current = voltage / resistance.In
symbols this is I = V / R. So our power
relationship power = V * I can be written
power = V * V / R = V^2 / R.
Using this relationship we find that
V = sqrt(power * R), so in this case the maximum voltage
(which will produce the 1/4 watt maximum power) will be
V = sqrt(1/4 watt * 2.7 * 10^3 ohms) = 26 volts.
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Question:
`qQuery general college
physics problem 18.39; compare power loss if 520 kW delivered at 50kV as
opposed to 12 kV thru 3 ohm resistance.** The current will not be the same at
both voltages.
Your Solution:
Confidence rating:
Given Solution:
It is important to understand that power (J / s) is the
product of current (C / s) and voltage (J / C).
So the current at 50 kV kW will be less than 1/4 the current
at 12 kV.
To deliver 520 kW = 520,000 J / s at 50 kV = 50,000 J / C
requires current I = 520,000 J/s / (50,000 J/C) = 10.4 amps. This demonstrates the meaning of the formula
P = I V.
To deliver 520 kW = 520,000 J / s at 12 kV = 12,000 J / C
requires current I = 520,000 J/s / (12,000 J/C) = 43.3 amps.
The voltage drops through the 3 Ohm resistance will be
calculated as the product of the current
and the resistance, V = I * R:
The 10.4 amp current will result in a voltage drop of 10.4
amp * 3 ohms = 31.2 volts.
The 43.3 amp current will result in a voltage drop of 40.3
amp * 3 ohms = 130 volts.
The power loss through the transmission wire is the product
of the voltage ( J / C ) and the current (J / S) so we obtain power losses as
follows:
At 520 kV the power loss is 31.2 J / C * 10.4 C / s = 325
watts, approx. At 12 kV the power loss
is 130 J / C * 43.3 C / s = 6500 watts, approx.
Note that the power loss in the transmission wire is not
equal to the power delivered by the circuit, which is lost through a number of
parallel connections to individual homes, businesses, etc..
The entire analysis can be done by simple formulas but
without completely understanding the meaning of voltage, current, resistance,
power and their relationships it is very easy to get the wrong quantities in
the wrong places, and especially to confuse the power delivered with the power
loss.
The analysis boils down to this:
I = P / V, where P is the power delivered. Ploss = I^2 R, where R is the resistance of
the circuit and Ploss is the power loss of the circuit.
So Ploss = I^2 * R = (P/V)^2 * R = P^2 * R / V^2.
This shows that power loss across a fixed resistance is
inversely proportional to square of the voltage. So that the final voltage, which is less than
1/4 the original voltage, implies more than 16 times the power loss.
A quicker solution through proportionalities:
For any given resistance power loss is proportional to the
square of the current.
For given power delivery current is inversely proportional
to voltage.
So power loss is proportional to the inverse square of the
voltage.
In this case the voltage ratio is 50 kV / (12 kV) = 4.17
approx., so the ratio of power losses is about 1 / 4.17^2 = 1 / 16.5 =
.06.
Note that this is the same approximate ratio you would get if you divided your 324.5 watts by 5624.7 watts. **
STUDENT COMMENT
This is confusing. I will have to keep reading over the
problem until I understand the
solution….I’m afraid that I will have problems with this test…we will see…….
INSTRUCTOR RESPONSE
The concept of thinking these ideas through in terms of the units is developed in the introductory problem sets, and may be helpful. A key idea here is that power (which you should understand is energy / time interval, measured in Joules/second or watts) is the product of current (in C / sec or amps) and voltage (in J / C or volts). If you multiply the number of Coulombs per second, by the number of Joules per Coulomb, you get the number of Joules per second.
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Question:
`qQuery univ students with 12th and later editions should attempt this problem
before reading the solution; 11th edition 25.62 (26.50 10th edition) A rectangular block
of a homogeneous material (i.e., with constant resistivity throughout) has
dimensions d x
2d x 3d. We have a source of potential difference V.
To which faces of the solid should the voltage be applied to attain
maximum current density and what is the density?
Your Solution:
Confidence rating:
Given Solution:
** First note that the current I is different for diferent
faces.
The resistance of the block is proportional to the distance
between faces and inversely proportional to the area, so current is
proportional to the area and inversely proportional to the distance between
faces. Current density is proportional
to current and inversely proportional to the area of the face, so current
density is proportional to area and inversely proportional to the distance
between faces and to area, leaving current inversely proportional to distance
between faces.
For the faces measuring d x 2d we have resistance R = rho *
L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d so current is I = V / R = V / (3/2
rho / d) = 2d V / (3 rho).
Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V
/ (3 rho d) = 1/3 V / (rho d).
For the faces measuring d x 3d we have resistance R = rho *
L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d so current is I = V / R = V / (2/3
rho / d) = 3 d V / (2 rho).
Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V
/ (2 rho d) = 1/2 V / (rho d).
For the faces measuring 3d x 2d we have resistance R = rho *
L / A = rho * (d) / (6 d^2) = 1 / 6 rho / d so current is I = V / R = V / (1/6
rho / d) = 6 d V / (rho).
Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V /
(rho d) = V / (rho d).
Max current density therefore occurs when the voltage is applied to the largest face. **
STUDENT QUESTION
I don’t really understand why current isn’t constant throughout the object. I can follow the solution, but I don’t understand that one point. My solution was based on the constant current which I see is not correct.
INSTRUCTOR RESPONSE
If the voltage is applied to different faces then you will
get different currents.
The potential gradient is greater if the faces are closer together. This effect
would tend to result in greater current for faces that are closer together.
If the faces are closer together the areas of the faces will be greater, so the
cross-sectional area will be greater. This effect would result in greater
current.
So the faces closest to together will experience the greatest potential
gradient, and result in the greatest cross-sectional area, resulting in the
greatest current.
However the question asks about current density, not current.
Intuitively, we know that the potential gradient is the electric field
responsible for accelerating the charges, so that the charge density will just
be proportional to the potential gradient, so that if the voltage is applied to
the faces with the least separation the current density will be the greatest. In
fact, since the separations are d, 2d and 3d we can see that the potential
gradients will be V / d, 1/2 V / d and 1/3 V / d so the other two current
densities will be 1/2 and 1/3 as great as the maximum.
This can be worked out symbolically for the general case:
R = rho * L / A,
and current is
I = V / R = V * A / (rho L)
so current density is
I / A = V / (rho L)
The greatest current density occurs for the least value of L, which for the
given situation is d, the separation between the largest faces.
Working out the details more fully is be unnecessary but might be instructive:
The faces closest together are separated by distance d, and have cross-sectional
area 2 d * 3 d = 6 d^2. So the resistance is
R = rho * L / A = rho * d / (6 d^2) = rho / (6 d) = 1/6 * rho / d
The faces furthest apart are separated by distance 3d, and have cross-sectional
area d * 2 d = 2 d^2. So the resistance is
R = rho * L / A = rho * d / (2 d^2) = rho / (2 d) = 1/2 * rho / d
Similar analysis shows that the resistance between faces separated by 2 d is rho
/ (3/2 d) = 2/3 * rho / d.
Thus the currents would be
I = V / R = 6 V * d / rho for the closest faces
I = V / R = 2 V d / rho for the most widely separate faces
and
I = V / R = 2 /3 V d / rho for the in-between separation.
The respective current densities, current / area, would be
(6 V d / rho) / (6 d^2) = V / (rho d)
(2 V d / rho) / (3 d^2) = 2/3 V / (rho d)
(2/3 V d / rho) / (2d^2) = 1/3 V / (rho d).
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