course Phy 232
7/19 1400
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?
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Self-critique (if necessary):
I am not in those classes.
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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?
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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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Okay.
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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?
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Self-critique (if necessary):
The magnetic field is 2*k*I/r at any point r.
The area of the slip is L*dr.
The magnetic flux through the slip is therefore 2*k*I*L*dr/r.
The total magnetic flux would be the integral from a to b of 2*k*I*L*dr/r or
2*k*I*L*ln(b/a).
If I is changing we get emf=2*k*L*ln(b/a)*di/dt.
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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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