course Phy 232
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.
031. `Query 31
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Question: `qQuery Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
Solution:
I’m in university physics
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Question: `q`qThe average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is
flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2.
The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of
flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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Question: `qquery gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output?
How many cycles per second are required to produce a 120-volt output, and how did you get your result?
Solution:
I’m in university physics
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Question: `q`qThe average magnitude of the output is peak output/sqrt(2) . We find the average output as ave rate of flux change.
The area of a single coil is (21 cm)^2 = (.21 m)^2 and the magnetic field is .65 Tesla; there are 320 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of
fluxMax = .65 T * (.21 m)^2 * 320 = 19.2 T m^2.
The flux will decrease to zero in 1/4 cycle. Letting t_cycle stand for the time of a complete cycle we have
ave magnitude of field = magnitude of change in flux / change in t = 9.17T m^2 / (1/4 t_cycle) = 36.7 T m^2 / t_cycle.
If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have
36.7 T m^2 / t_cycle = 120 V / sqrt(2).
We easily solve for t_cycle to obtain t_cycle = 36.7 T m^2 / (120 V / sqrt(2) ) = .432 second.+
A purely symbolic solution uses
maximum flux = n * B * A
average voltage = V_peak / sqrt(2), where V_peak is the peak voltage
giving us
ave rate of change of flux = average voltage so that
n B * A / (1/4 t_cycle) = V_peak / sqrt(2), which we solve for t_cycle to get
t_cycle = 4 n B A * sqrt(2) / V_peak = 4 * 320 * .65 T * (.21 m)^2 * sqrt(2) / (120 V) = .432 second.
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Question: `quniv query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire.
When the current is I what is the magnitude of B at distance r from the wire and what is the magnetic flux through a strip at this position having width `dr?
Solution:
A = L*dr
Magnetic flux = 2*k*I/r*(L*dr)
Total magnetic field = sum(s*k*I/r*L*dr) r goes from a to b
Using the limit as dr goes to 0, take the integral of:
2*k*I/r*L dr, going from a to b
Doing the antiderivative of this and then finding the definit integral,
Flux = 2*k*L*ln(b/a)*I
dflux/dt = 2*k*L*ln(b/a)*dI/dt
If,
a = 12.0 cm = 0.12 m
b = 36.0 cm = 0.36 m
L = 24.0 cm = 0.24 m
dI/dt = 9.60 A / s
k = 9*10^9 N*m^2/C^2
dflux/dt = 4.56*10^10
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Question: `q`q** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page.
The area of the strip is L * `dr.
The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr).
The total magnetic field over a series of such strips partitioning the area is thus
sum(2 k ' I / r * L `dr, r from a to b).
Taking the limit as `dr -> 0 we get
integral (2 k ' I / r * L with respect to r, r from a to b).
Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to
flux = 2 k ' L ln | b / a | * I.
If I is changing then we have
rate of change of flux = 2 k ' L ln | b / a | * dI/dt.
This is the induced emf through a single turn.
You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. **
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