Respuesta :
Answer;
= 0.2858 N, attractive
Explanation;
Since the currents are parallel, the force on each wire will be attractive, toward the other wire. Therefore; to calculate the magnitude of force we use the formula;
F2 = (μ₀/2π)(I₁I₂/d)I₂
Where;
μ₀ = constant = 4π × 10^-7 H/m,
I₁, I₂ are the currents,
L s the length o the wire,
d is the distance between these two wires
if current are in same direction, they exert attractive force on each other and if currents are in opposite direction, they exert repulsive force
Therefore;
F₂ = (4π × 10^-7 T.m/A)/2π) × ((35 A)²/ 0.03 m)× 35 m
= 0.2858 N, attractive
The magnitude and direction of the force between two parallel wires will be 0.2858 N, attractive
What will be the magnitude and direction of the force between two parallel wires?
It is given that the currents are flowing parallel in the wire so the force on each wire will be attractive. By using the force formula for two parallel wires carrying current.
[tex]F=\dfrac{\mu}{2\pi } \dfrac{I_1 I_2}{d} L[/tex]
Where:
μ₀ = constant = [tex]4\pi \times10^{-7} \ \frac{H}{m}[/tex]
I₁, I₂ are the currents,
L = is the length o the wire,
d = is the distance between these two wires
if the current is in the same direction, they exert an attractive force on each other and if currents are in opposite direction, they exert a repulsive force
Therefore
[tex]F=\dfrac{4\pi\times10^{-7}}{2\pi} \dfrac{35\times35}{0.03} \times35[/tex]
[tex]F=0.2858N[/tex]
Thus the magnitude and direction of the force between two parallel wires will be 0.2858 N, attractive
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