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Wittaler [7]
3 years ago
8

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

2=0.75 Amps. They are 5 meters long in this configuration and separated from each other by 5 cm. What is the magnitude of the force experienced by the wire carried I2?
Physics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:F=3\times 10^{-5} N

Explanation:

Given

I_1=2 Amps

I_2=0.75 Amps

Length of each wires\left ( L\right )=5 m

Distance between wires \left ( r\right )=5 cm

Force per unit length =\frac{\mu_0I_1I_2}{2\pi r}

F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}

F'=6\times 10^{-6}

Force for L=5 m

F=3\times 10^{-5} N

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Answer:

W=76.55 miles.metric tons

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W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx

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A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670
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Answer:  459.14 N

Explanation:

from the question, we have

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∑ F = F c =  F w − Fn ..............equation 1

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substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2)   ..........equation 3

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therefore m = Fw / g

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Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

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