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

c because the news are a source of information of things that happen on the daily and shows evidence about that particular topic

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west