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Bogdan [553]
3 years ago
13

A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co

nnected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole mo- ment of this device? (b) At what axial distance d will the mag- netic field have the magnitude 5.0 mT (approximately one-tenth that of Earth’s magnetic field)?
Physics
1 answer:
kupik [55]3 years ago
3 0

Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

Where;

N is number of turns of coil

I is current in wire

A is area

We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

μ = 300 x 4 x 0.001963 = 2.36 A.m².

B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

Thus,

z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)

Solving this gives; z = 0.46m =

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Answer:

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Explanation:

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v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\

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Substituting

v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=337.31m/s \\

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v_{air}=337.31\times 3.28=1106.38ft/s \\

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Sound level B in decibels is defined as
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Answer:

The approximate combined sound  intensity is I_{T}=1.1\times10^{-4}W/m^{2}

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

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In the same way for the loud conversation having a decibel intensity of 70 dB.

10log(\frac{I_{2}}{I_{0}} )=70, therefore I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}

Finally we add both of them I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}, is the approximate combined sound  intensity.

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A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r
katen-ka-za [31]

Answer:

  • k = 167.33 N/m

Explanation:

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  • v = c = λ*f

        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
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