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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 −
Question
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.
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Physics
4 years
2021-07-27T17:51:15+00:00
2021-07-27T17:51:15+00:00 2 Answers
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Answers ( )
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
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