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stealth61 [152]
2 years ago
15

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

e that produces a 0.10T magnetic field. Note: No = 4 TX10-7T-m/A and copper = 1.72 x 10-82.m. Note: Do not assume the voltage transmitted on the power line. Note: This is a two-step problem: This is a two-step problem: Calculate the radius of the power line from the physical dimensions and then calculate the current from the magnetic field. A. 125A OB. 250A C.500A OD. 750A E. 1,000A
Physics
1 answer:
Oliga [24]2 years ago
4 0

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

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air gas

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3 years ago
To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef
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Answer:

29.76245 rad/s², -117.80972 rad/s²

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Explanation:

\omega_f = Final angular velocity

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\theta = Angle of rotation

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Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2

Angular acceleration during speed up is 29.76245 rad/s²

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2

Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

\omega=2\pi 4.5=28.2743\ rad/s

Maximum angular speed reached by the flywheel is 28.2743 rad/s

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad

The ratio would be \dfrac{13.4303}{3.39292}=3.95833

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