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 total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)