0.00025 N is the magnitude of the net force on the first wire in (figure 1).
The total of all forces exerted on an item is known as the net force. A mass can accelerate due to net force. A body is subject to another force whether it is at rest or in motion. When there are a lot of forces acting on a system, the phrase "net force" is employed.
From the diagram, first wire has an attracting force from third wire at 0.04 m distance and a repelling force from second wire at 0.02 m distance.
Expression of the force is -
F = [µ ×π×(I)² × L] / (2πd)
When d = 0.02 m,
or, F = [4 ×π× 10⁻⁷(10)² × 0.50] / (2π × 0.02)
or, F = 0.0005 N
And when d = 0.04 m,
or, F = [4 ×π× 10⁻⁷(10)² × 0.50] / (2π × 0.04)
or, F = 0.00025 N
So, the net force on first wire = 0.005 - 0.00025
= 0.00025 N repelling or upward.
To know more about net force refer to:
https://brainly.com/question/12970081
#SPJ1
The complete question is as follows:
What is the magnitude of the net force on the first wire in (Figure 1)?
Express your answer in newtons.
