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slamgirl [31]
2 years ago
12

How many turns should a 10-cm long ideal solenoid have if it is to generate a 1.5-mT magnetic field when 1.0 A of current runs t

hrough it? (μ0 = 4???? × 10-7 T · m/A)
Physics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

Answer:

N=119.34 turns

Explanation:

The magnetic field of a solenoid is calculated using the formula:

B= µo*\frac{I*N}{L} Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=10cm=10*10^{-2}, B= 1.5mT=1.5*10^{-3} T  ,I=1A  

µo=4\pi *10^{-7} \frac{T*m}{A}

We cleared N of the equation (1):

N=B*L/ µo*I

N= (1.5*10^{-3} *10*10^{-2} )/(4\pi *10^{-7} *1

N=0.1193*10^{3}

Answer

N=119.34 turns

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Answer:

<h2>5.6kW</h2>

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2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

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7 0
2 years ago
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Misha Larkins [42]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2 
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The table shows the decay in a 59 g sample of bismuth 212 overtime.
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Answer: C

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