>course Phy 202
>This one was way shorter than the others. Did I do the whole
thing? It ended on 'subscript error'.I meant to ask on the
second problem, this could be and
>probably is very elementary, but how does the average voltage
come from the maximum over sqrt(2)? Where does the square root
of 2 come from? Thanks!
>|??P????????x??assignment #010
>
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>
>010. `Query 31
>Physics II
>10-12-2007
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> 07:43:23
>Query 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?
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>RESPONSE -->
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> 07:43:26
>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.{}{}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
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>RESPONSE -->
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> 07:43:29
>flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec
= .053 volts.
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>RESPONSE -->
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> 07:46:05
>query 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?
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>RESPONSE -->
>I worked on this problem for a while, and I could be
algebraically challeneged. It looks like the formula:
>emf = N*B*angular velocity*area*sin(angular velocity * time),
but I was unable to get angular velocity on its own.
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> 06:34:07
>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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>RESPONSE -->
>ave rate of flux change = peak output / sqrt (2)
>A = .0441 m^2, B = .65 T, N = 320. Theta = 90 degrees
>flluxMax = N*A*B = 320 * .0441 m^2 * .65 T = 19.2 Tm^2
>After 1/4 cycle, flux = zero.
>B_avg = change in flux / change in time
>B_avg = 19.2 Tm^2 / (1/4 t) = 36.7 Tm^2 / t
>V_avg = 120 V / sqrt(2)
>36.7 Tm^2 / t = 120 V/sqrt(2)
>t = 36.7 Tm^2 / (120 V / sqrt(2)) = .432 s
>.25 cycle / .432 second = .579 cycles/second produces the
maximum output of 120 Volts.
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> 06:34:11
>univ 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?
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>RESPONSE -->
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> 06:34:14
>** 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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>"
>end of document
>Your work has not been reviewed.
>Please notify your instructor of the error, using the Submit
Work form, and be sure to include the date 10-15-2007.
>This looks good.
&#Let me know if you have questions. &#