In a classroom demonstration, two long parallel wires are separated by a distance of 2.25 cm. One wire carries a current of 1.30 A while the other carries a current of 3.15 A. The currents are in the same direction. (a) What is the magnitude of the force per unit length (in N/m) that one wire exerts on the other

Explanation:

Given that,

Current in wire are 1.3A and 3.15A

Distance between wire is d= 2.25cm

d = 2.25/100 = 0.025m

Force per unit length F/l?

Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2).

The field due to I1 at a distance r is given to be

B1 = μo• I1 / 2πr

This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by

F=ILBsinθ

with sinθ=1:

F2=I2 • L •B1

By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2. (Note that F1=−F2.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for B1 into the last equation and rearranging terms gives

F/l = μo• I1• I2 / 2πr

Where μo is constant

μo = 4π×10^7 Tm/A

Then,

F/l = μo• I1• I2 / 2πr

F/l = 4π ×10^-7 × 1.3×3.15/(2π×0.025)

F/l = 3.276×10^-5 N/m

the magnitude of the force per unit length that one wire exerts on the other is 3.276×10^-5 N/m

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.