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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
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?
Your Solution:
Confidence rating:
Given Solution:
The 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.
Flux is designated by the Greek letter phi.
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
`d phi / `dt = .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
STUDENT COMMENT
OK so its in Volts. I understand
INSTRUCTOR RESPONSE
You had the right number. You should also carry the
units throughout the calculation.
A Tesla is a N / (amp m) so the unit T m^2 / sec becomes N m / (amp sec) = J /
(C/s * s) = J / C, or 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?
Your Solution:
Confidence rating:
Given Solution:
The 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?
Your Solution:
Confidence rating:
Given Solution:
** 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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Question: A 320-loop square coil 21 cm on a side
rotates about an axis perpendicular to a .65 T mag field. What frequency of
oscillation will produce a peak 120-v output?
Your Solution:
confidence rating:
Given Solution:
I wouldn't advocate using a formula from the book to solve
this problem. Common sense, starting from the premise that the voltage function
is the derivative of the flux function, is much more efficient (way fewer
formulas to remember and less chance of using the wrong one).
The maximum flux is .65 T * (.21 m)^2/ loop * 320 loops = 9.173 T m^2.
So the flux as a function of clock time could be modeled by
phi(t) = 9.173 T m^2 * sin(2 pi f t).
The voltage induced by changing flux is the rate of change
of flux with respect to clock time. So the
voltage function is the t derivative of the flux:
V(t) = phi ' (t) = 9.173 T m^2 * 2 pi f cos(2 pi f t).
Maximum voltage occurs when cos(2 pi f t) = 1. At this instant the voltage is
max voltage = 9.173 T m^2 * 2 pi f.
Setting this equal to the peak voltage we get
9.173 T m^2 * 2 pi f = 120 V so that
f = 120 V / (9.173 T m^2 * 2 pi) = 2.08 V / / (T m^2) = 2.08 T m^2 / s / (T m^2) = 2.08 s^-1.
We can generalize this symbolically by replacing 9.173 T
m^2 by phi_max, which represents the maximum
flux. So a generator with maximum flux phi_max, rotating a frequency f has flux
function
phi(t) = phi_max cos(2 pi f t) with t derivative
V(t) = phi ' (t) = phi_max cos(2 pi f t).
Everything follows easily from this formulation, with no
need to memorize the formulas that result.
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