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3 years ago

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Two long parallel wires carry currents of 1.77 A and 8.93 A . The magnitude of the force per unit length acting on each wire is

5.93×10−5 N/m . Find the separation distance of the wires expressed in millimeters.

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

533mm

Explanation:

The concept that we need to apply here to find the radius is the Force in a Magnetic Field,

That is basically Force per unit length in terms of the current, which is given by,

$F= \frac{\mu_0I_1I_2}{2\pi R}$

Where,

$\mu_0$ is the permeability constant

$I_i$ is the current

R the radius

Re-arrange for Radius,

$R = \frac{\mu_0 I_1I_2}{2\pi R}$

Replacing our values

$R = \frac{4\pi*10^{-7}(1.77)(8.93)}{5.93*10^{-5}2\pi}$

$R = 0.0533m$

Therefore the separation distance of the wires is 533mm

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Answer:

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Answer:

$x=22.65m$

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Answer:

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We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

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Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

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$G^2 (1+\frac{f}{f_c})^{2n}= 1$

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$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

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$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

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And replacing we got:

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