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zalisa [80]
3 years ago
13

Critical thinking requires that _______. a. conclusions be held constant after their initial determination b. conclusions be adj

usted as necessary to incorporate new knowledge c. information from a particular source only be verified the first time the source is used d. there be a single solution to a particular problem Please select the best answer from the choices provided
Physics
2 answers:
kykrilka [37]3 years ago
8 0

Answer

its B..................................       just took the test

Explanation:

g100num [7]3 years ago
7 0

a. conclusions be held constant after their initial determination

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A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10-4 (T·m
shepuryov [24]

Answer:

(a)The current passes through the solenoid is 11.7 A.

(b) The new current will be one-fourth of the initial current.

Explanation:

Given that,

The number of turns per meter = 385

Diameter of solenoid = 17.0 cm = 17\times 10^{-2} m

Magnetic flux  through core of solenoid \phi = 1.28\times 10^{-4} Tm²

(a)

Magnetic field B= \mu_0nI

\mu_0= 4\pi \times 10^{-7} T/amp m

Cross section area of the solenoid A= \pi \frac{d^2}{4}

                                                             =\pi\frac{ (17\times 10^{-2})^2}{4}  m²

The angle between magnetic field and cross section of the solenoid is \theta =0^\circ

The magnetic flux through a area A with magnetic fie;d B is

\phi = BA cos\theta

\Rightarrow \phi =( \mu_0nI)(\pi \frac{d^2}4)cos \theta

\Rightarrow I =\frac{\phi}{(\mu_0\pi n \frac{d^2}4cos\theta)}

       =\frac{4\phi}{(\mu_0n)(\pi d^2)cos\theta}

      =\frac{1.28\times 10^{-4}\times 4}{(4\pi \times10^{-7}\times385 )\times(\pi\times17\times 10^{-2})^2cos 0^\circ}

     =11.7 A

The current of the solenoid is 11.7 A.

(b)

I =\frac{4\phi}{(\mu_0\pi n d^2cos\theta)}

From the above equation it is clear that, the current is inversely proportional to the square of the diameter of a solenoid.

I\propto \frac1{d^2}.

Consider d' be the new diameter of the solenoid .

Since the new diameter of the solenoid is double of the initial diameter.

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\frac{I}{I'}= \frac{(d')^2}{d^2}

\Rightarrow \frac{I}{I'}=\frac{(2d)^2}{d^2}

\Rightarrow \frac{I}{I'}=4

⇒I=4I'

\Rightarrow I'=\frac{I}{4}

The new current will be one-fourth of the initial current.

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