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sdas [7]
3 years ago
10

Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation

E(t)=−Ldi(t)/dt. Find L for a long solenoid with n turns. (Hint: The self-inductance L will always be a positive quantity.), Epxress in μ0, di, dt, R and n
Physics
1 answer:
Artemon [7]3 years ago
4 0

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

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klemol [59]

Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

Metalloids have properties that straddles between those of metals and non-metals.

In essence, they can be conductive or not under certain conditions.

The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

3 0
3 years ago
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How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find
vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, a_c = v²/r

∴ The centripetal acceleration, a_c = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, a_c = 20·π/9 m/s² ≈ 6.98 m/s²

4 0
2 years ago
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga
Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

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