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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Answer:
The Border Patrol of the United States analyzes the purchase of a new helicopter for the aerial surveillance of the border of New Mexico and Texas with the Mexican Republic. 4 years ago a similar helicopter was purchased at a cost of $ 140,000.00. with an interest rate of 7% per year. Calculate the single payment factor and the present value factor with the above data with the table and formula. Draw the flow chart.
Explanation:
This question involves the concepts of power and current.
The power output of the motor is "10.08 hp".
<h3>Power Output</h3>
The power output of the motor can be found using the following formula:
P = ηVI
where,
- P = Power = ?
- η = efficiency = 90 % = 0.9
- I = current = 38 A
- V = potential difference = 220 V
Therefore,
P = (0.9)(38 A)(220 V)
P = 7524 watt
converting this to horsepower:
P = (7524 watt)()
P = 10.08 hp
Learn more about power here:
brainly.com/question/13948282
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