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2 years ago

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If the volume of a container of gas remains constant, what will happen to the pressure of a gas if you increase the temperature?

1 answer:

Scorpion4ik [409]2 years ago

4 0

Answer:

This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure.

Explanation:

Hope it helps

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4 years ago

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

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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.

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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3 years ago