Answer:
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
