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3 years ago

11

You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you

r wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 570 turns. The diameter of the tube is 8.10 cm, and the length of the wire-wrapped portion is 35.0 cm. You pull out your calculator to determine the following.
a. the inductance of the coil (in mH)
b. the emf generated in the coil if the current in the wire increases at the rate of 3.00 A/s (Enter the magnitude in mV.)

1 answer:

telo118 [61]2 years ago

7 0

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l = 35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L = 0.00841549 / 1.4

L = 6 × 10⁻³ H

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV

Therefore, the emf generated in the coil is 18 mV

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3 years ago

A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the

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Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I) ⇒ V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒ Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

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3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

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Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago