Physics, 30.03.2020 18:01 kkeith121p6ujlt

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·m2 ). (a) Find the current in this solenoid. (b) How would your answer in part (a) change if the diameter of the solenoid were doubled? Explain your detailed reasoning and give the answer.

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Answer from: jholbrook7643

A) The current passing through the solenoid is = 11.7 Amperes

B) When the diameter of the solenoid is doubled the current = 2.93 A

Given data :

Number of turns ( n ) = 385

Diameter of solenoid ( d ) = 17.0 cm = 17 * 10⁻² m

Magnitude of magnetic flux ( ∅ ) = 1.28 * 10⁻⁴ T.m²

A) Determine the current in the solenoid

Current through the solenoid ( I ) = [ 4∅ / ( μο*n )*(πd²) *cosθ ) ] --- ( 1 )

= [ ( 4 * 1.28 * 10⁻⁴ ) / ( 4π * 10⁻⁷ * 385 ) * ( π * 17 * 10⁻² )² * cosθ ) ]

= 11.7 Amperes

B) when the diameter is doubled

( I ) = [ 4∅ / ( μο*n )*(πd²) *cosθ ) ] ( 2 )

From the equation above

I ∝ 1/d² ( i.e. current in solenoid is inversely proportional to the diameter )

∴ when the diameter of the solenoid is doubled

The value of the new current ( I' ) = 1/4 * ( I )

= 1/4 * 11.7

≈ 2.93 amperes.

Hence we can conclude that The current passing through the solenoid is = 11.7 Amperes, When the diameter of the solenoid is doubled the current will be 2.93 A

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Answer from: marissaa190

(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.

That is d'= 2d.

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

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

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

⇒I=4I'

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

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

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A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core o...