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irina [24]
2 years ago
12

How many turns should a 10-cm long solenoid have if it is to generate a 1.5 x 10-3 t magnetic field on 1.0 a of current?

Physics
1 answer:
stepan [7]2 years ago
3 0

We can answer this problem using Ampere’s Law:

<span>Bh = μoNI </span>

Where:

B = Magnetic Field

h = coil length

<span>μo = permeability =4π*10^-7 T·m/A </span>

N = number of turns

I = current

It is given that B=0.0015T, I=1.0A, h=10 cm = 0.1m<span>

Use Ampere's law to find # turns: 
Which can be rewritten as: 
<span>N = Bh/μoI </span>
N = (0.0015)(0.1)/(4π*10^-7)(1.0) 
N = 119.4 

</span>

<span>Answer: 119.4 turns</span>

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2 years ago
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Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

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T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

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H_a=1491*1.25=1863.75W

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