Assignment 31

course Phy 202

8/4 5:30p

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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Your Solution:

Flux= Magnetic field* area

Flux= (1.10 T)*(pi* .048 m)^2

Flux= .00796 T m^2

Rate of change of flux= .00796 T m^2/ (.15 sec)

Rate= .053 T m^2/sec

confidence rating:

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2

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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.

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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Self-critique (if necessary):

ok

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Self-critique Rating:3

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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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Your Solution:

Avg magnitude= peak output/`sqrt(2)

Avg output= avg rate of flux change

Max flux= magnetic field* area*# of coils

Max flux= (.65 T)*(.21 m)^2*(320)

Max flux= 19.2 T m^2

Avg magnitude= mag. Of change in flux/ `dT

Avg mag= (9.17T m^2)/(1/4 cycle)

Avg mag= 36.7 T m^2/ t_cycle

Avg voltage= 120 V/sqrt(2), therefore

36.7 T m^2/t_cycle= 120 V/ sqrt(2)

Solve for t_cycle

T_cycle= 36.7 T m^2 / (120 V / sqrt(2) ) = .432 second

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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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Self-critique (if necessary):

ok

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Self-critique Rating:

3

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