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irakobra [83]
3 years ago
8

A wire with a weight per unit length of 0.082 N/m is suspended directly above a second wire. The top wire carries a current of 3

0.6 A and the bottom wire carries a current of 60.4 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. mm
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0

Answer:

d = 4.5079mm

Explanation:

from the question we are given that

\frac{F}{L}= 0.082N/m

I₁= 30.6 A , I₂ = 60.4 A , and μ₀ = 4 π × 10⁻⁷ T · m / A

In order for the system to be in equilibrium, the repulsive magnetic force

per unit length on the top wire must equal the weight per unit length of the wire. Then we have

F/L = μ₀I₁I₂ / 2πd , making d as the subject of formular

we have that d = Lμ₀I₁I₂ / 2πF

= 4 π × 10⁻⁷ × 30.6 ×60.4 / 2π × 0.082

=45079.02×  10⁻⁷

in mm .. 45079.02×  10⁻⁷ × 10³

45079.02 ×10⁻⁴

d = 4.5079mm

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let's replace

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4 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
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Answer:

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Explanation:

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<span>So the upward force is ≈ </span><span>0.45N. </span>
6 0
3 years ago
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