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3 years ago

a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of

Physics

1 answer:

Natalija [7]3 years ago

8 0

Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Answer:

(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T$

(a) Calculate the energy density of the magnetic field inside the solenoid

$u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3$

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴ x 0.986)

U = 0.121 J

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Answer:

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So, substituting the variables into the equation, we have

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Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

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3 years ago

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Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

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we know that;

shear stress τ = 16T/πd³

where d is the diameter

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

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now from the second image;

lets length dx which is at distance of "x" from "B"

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T(x) = tx

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d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

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3 years ago

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Answer:

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$\Delta V=Ed$

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