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rosijanka [135]
3 years ago
15

Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is

6.03 × 10 − 5 N / m . Find the separation distance d of the wires expressed in millimeters.
Physics
2 answers:
Marysya12 [62]3 years ago
8 0

Answer:

d = 77.5mm

Explanation:

The expression used for calculating the force between two parallel wires carrying currents separated by a distance is;

F12 =µµol1I2L/2πd where

I1 and I2 are the currents in both wires

L is the length of the wires

d is the separation distance

F12 is the magnitude of the force between them.

µo is the constant of proportionality

The equation can be rewritten as;

F12/L = µµol1I2/2πd

F12/L is the force per unit length acting on each wire= 6.03×10^-5 N/m

I1 = 3.35A

I2 = 6.99A

µ = 1

µo = 4π×10^-7N/A²

d =?

Substituting this datas in the equation above to get the separation distance d we have;

6.03×10^-5 =1× 4π×10^-7 × 3.35 × 6.99/2πd

6.03×10^-5 = 4×10^-7×23.42/2d

6.03×10^-5 = 9.37×10^-6/2d

2d = 9.37×10^-6/6.03×10^-5

2d = 1.55 × 10^-1

d = (1.55 × 10^-1)/2

d =0.155/2

d = 0.0775m

Since 1m = 1000mm

0.0775m = (0.0775×1000)mm

d = 77.5mm

Nady [450]3 years ago
7 0

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

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So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
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The work done by a moving object can be expressed in the equation:

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\boxed{\sf{\bold{W = F \times s}}}

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\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

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<h3>Solution</h3>

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  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

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